LIMITS

In this chapter, you should learn and meet the simulate pollutant dispersion and climate shifts,
following objectives: aiding disaster preparedness. Medicine
leverages limits to determine steady-state drug
Illustrate the limit of a function using a concentrations, refining treatment protocols.
table of values and the graph of the Technology advances through limit-based
function. analysis of natural processes, enhancing system
performance. Statistics uses the Central Limit
Distinguish between 𝑓(𝑥) and 𝑓(𝑐). Theorem to simplify data and inform decisions,
from elections to industrial quality control.
Illustrate the limit theorems. Population studies analyze growth trends and
equilibrium states via limits. By modeling
Apply the limit theorems in evaluating behavior near critical points or infinity, limits
the limit of algebraic functions provide a framework to solve real-world
(polynomial, rational, and radical); problems, ensuring safety, efficiency, and
innovation across industries.
Evaluate infinite limits, limits at infinity,
and infinite limits at infinity.

Compute the limits of exponential,

logarithmic, and trigonometric functions
using a table of values and graphs of
the functions.

Introduction

In mathematics, a limit is a fundamental

concept that describes the value a function or
Limits are crucial for analyzing and improving the performance of
sequence approaches as the input or index gets
designs, ensuring structural integrity, and optimizing systems.
closer to a specific point or infinity. Limits are Engineers use limits to determine maximum load capacities, optimize
essential in calculus and mathematical analysis fluid flow, and enhance digital communication systems​
​
and are used to define key concepts such as
derivatives, integrals, and continuity. They help Limit Theorem
analyze the behavior of functions near specific
points, enabling precise calculations and The limit theorem refers to a set of
predictions.Limits are foundational in real-world mathematical principles that describe how limits
applications across disciplines, enabling precise behave under various operations and conditions.
modeling and optimization of phenomena. In These theorems are foundational in calculus and
engineering, they determine structural load mathematical analysis, providing tools to
limits and optimize systems like fluid dynamics evaluate limits of functions or sequences.
or digital communication. Physics relies on
limits to analyze motion, fields, and energy
transfer, while chemistry uses them to predict lim 𝑓(𝑥) = 𝑎
reaction outcomes and temperature changes. 𝑥→𝑐
Economics applies limits to optimize costs,
forecast trends, and calculate marginal values.
Environmental science employs limits to
1
Limit of a Constant 5.​ lim 𝑥 = 99
𝑥 → 99

The limit of a constant is the constant

itself. In mathematical terms, if c is a constant (a
fixed numerical value)
Limit of a Linear function
lim 𝑓(𝑥) = 𝑎
𝑥→𝑎 The limit of a linear function refers to
where 𝑐 is a constant and 𝑎 is any real number
evaluating the behavior of a function of the form
f(x)=mx+b, where m and b are constants, as x
SAMPLE PROBLEMS | Limit of a Constant approaches a specific value or infinity.

1.​ lim 7 = 7
𝑥→3
lim 𝑐𝑥 = 𝑐 lim 𝑥
𝑥→𝑎 𝑥→𝑎
Where c and a are real numbers
2.​ lim 4 = 4
𝑥→4
SAMPLE PROBLEMS | Limit of a Linear
3.​ lim − 5 = -5 Equation
𝑥→3

1.​ lim 2𝑥 = 2 lim 𝑥

4.​ lim 5 = 5 𝑥→4 𝑥→4
𝑥 → −6 = 2 (4)
=8
5.​ lim − 12 = 12
𝑥→6
2.​ lim − 5𝑥 = − 5 lim 𝑥
𝑥→3 𝑥→3
= -5 (3)
Limit of the Identity Function = -15

The limit of the identity function refers to 3.​ lim 12𝑥 = 12 lim 𝑥
𝑥→2 𝑥→2
evaluating the behavior of the function f(x) = x as
= 12 (2)
x approaches a specific value
= 24

lim 𝑥 = 𝑎 4.​ lim 36𝑥 = 36 lim 𝑥

𝑥→𝑎 𝑥 → −2 𝑥 → −2
where 𝑎 is any real number = 36 (-2)
= -72
SAMPLE PROBLEMS | Limit of the Identity
Function 5.​ lim − 20𝑥 = − 20 lim 𝑥
𝑥 → −4 𝑥 → −4

1.​ lim 𝑥 = 5 = -20(-4)

𝑥→5 = 80

2.​ lim 𝑥 = -5 Limit of a Power

𝑥 → −5

The limit of a power refers to evaluating

3.​ lim 𝑥 = 4
𝑥→4 the limit of a function raised to a power. If the
limit of a base function f(x) exists as x→a, then
4.​ lim 𝑥 = 11 the limit of its power can be computed by raising
𝑥 → 11
the limit of the base function to the same power.
2
 Limit of the Sum and Difference of
If n is a positive integer and lim 𝑓(𝑥) = 𝐿, then
𝑥→𝑎
Functions
𝑛 𝑛
lim (𝑓(𝑥)) = ( lim 𝑓(𝑥))
𝑥→𝑎 𝑥→𝑎 The limit of the sum or difference of two
𝑛 functions follows specific algebraic rules that
= 𝐿
allow us to break down complex expressions
into simpler parts.

If lim 𝑓(𝑥) = 𝐿 and lim 𝑔(𝑥) = 𝑀, then

𝑥→𝑎 𝑥→𝑎
SAMPLE PROBLEMS | Limit of a Power
lim [𝑓(𝑥) ± 𝑔(𝑥)] = lim 𝑓(𝑥) ± lim 𝑔(𝑥)
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎
2
1.​ lim (3) = ( lim 3)2
𝑥→0 𝑥→0
= 𝐿±𝑀
= (3)2
=9
SAMPLE PROBLEMS | Limit of the Sum and
2 Difference of Functions
2.​ lim (2𝑥) = ( lim 2𝑥)2
𝑥→3 𝑥→3
= (2 lim 𝑥)2 1.​ lim (2𝑥+ 5)
𝑥→3 𝑥→2
= (2(-2))-2 = lim 2𝑥 + lim 5
= (6))2 𝑥→2 𝑥→2

= 36 = 2 lim x + lim 5
𝑥→2 𝑥→2

2 =2(2) + 5
3.​ lim (4𝑥) = ( lim 4𝑥)2 =4 +5
𝑥 → −2 𝑥 → −2
=9
= (4 lim 𝑥)2
𝑥 → −2
2
= (4(-2))2 2.​ lim (𝑥 − 3)
= (-8)2 𝑥→2

= 64 = ( lim 𝑥)2 - lim 3

𝑥→2 𝑥→2

−2 =(2)2 -3
4.​ lim (2𝑥) = ( lim 2𝑥)-2 =4 -3
𝑥 → −2 𝑥 → −2
=1
= (2 lim 𝑥)-2
𝑥 → −2
2
= (2(-2))-2 3.​ lim (𝑥 + 2𝑥 − 4 )
= (-4)-2 𝑥 → −2
1 = ( lim 𝑥)2 + lim 2𝑥 - lim 4
= 2 𝑥 → −2 𝑥 → −2 𝑥 → −2
(−4)

=
1 =( lim 𝑥)2 + 2 lim 𝑥 - lim 4
16 𝑥 → −2 𝑥 → −2 𝑥 → −2
1 2 1 2
5.​ lim ( 2 𝑥) = ( lim 𝑥)2 =(-2) + 2(-2)-4
2
𝑥→3 𝑥→3 =4-4-4
1
= ( 2 lim 𝑥)2 =-4
𝑥→3
1
= ( 2 (3))2 4.​ lim ( 2 𝑥 −
1 1
)
4
3 2 𝑥→2
=( 2
) 1 1
= lim 2
𝑥 - lim 4
9
= 𝑥→2 𝑥→2
4
1 1
= 2
( lim 𝑥) - lim 4
𝑥→2 𝑥→2
1 1
= 2
(2) - 4

3
 =1-
1 3.​ lim (𝑥 + 2)(3)
4 𝑥→2
3
= 4 =( lim 𝑥 + lim 2)( lim 3)
𝑥→2 𝑥→2 𝑥→2

2 5
=(2+2)(3)
5.​ lim (3𝑥 + 6
𝑥) =(4)(3)
𝑥→2
2 5
=12
= ( lim 3𝑥) + ( lim 6
𝑥)
𝑥→2 𝑥→2
2 5
4.​ lim (𝑥 + 3)(𝑥 − 4)
= 3( lim 𝑥) + 6 ( lim 𝑥) 𝑥 → −1
𝑥→2 𝑥→2
=( lim 𝑥 + lim 3)(
5 𝑥 → −1 𝑥 → −1
=3(2)2+ ( 6 (2))
10
lim 𝑥 − lim 4)
=3(4)+ ( 6
) 𝑥 → −1 𝑥 → −1
10 =(-1 + 3)(-1 -4)
=12+ 6 =(2)(-5)
41
= 3 = - 10

5.​ lim (2𝑥 + 3)(4𝑥 − 1)

Limit of the Product of Functions 𝑥→2
​ =( lim 2𝑥 + lim 3)( lim 4𝑥 − lim 1)
The limit of the product of two functions 𝑥→2 𝑥→2 𝑥→2 𝑥→2
follows a straightforward rule in calculus, =(2 lim 𝑥 + lim 3)(4 lim 𝑥 − lim 1)
allowing us to decompose complex expressions 𝑥→2 𝑥→2 𝑥→2 𝑥→2

into simpler parts. =(2(2))+3)(4(2)-1)

=(4+3)(8-1)
=(7)(7)
If lim 𝑓(𝑥) = 𝐿 and lim 𝑔(𝑥) = 𝑀, then, =49
𝑥→𝑎 𝑥→𝑎

lim [𝑓(𝑥)𝑔(𝑥)] =
𝑥→𝑎
( lim 𝑓(𝑥)
𝑥→𝑎
)(
𝑥→𝑎
)
lim 𝑔(𝑥)

= 𝐿𝑀
Limit of the Quotient of Functions

The limit of the quotient of two functions

SAMPLE PROBLEMS | Limit of the Product of follows specific rules in calculus, enabling the
Functions evaluation of limits involving division.

1.​ lim (𝑥)(2) If lim 𝑓(𝑥) = 𝐿 and lim 𝑔(𝑥) = 𝑀, then

𝑥→3 𝑥→𝑎 𝑥→𝑎

= ( lim 𝑥) ( lim 2) lim 𝑓(𝑥)

𝑥→3
=(3)(2)
𝑥→3 lim
𝑥→𝑎
( 𝑓(𝑥)
𝑔(𝑥) )= 𝑥→𝑎
lim 𝑔(𝑥)
𝑥→𝑎
=6 𝐿
= 𝑀
𝑤ℎ𝑒𝑟𝑒 𝑚 ≠ 0
2.​ lim (2𝑥)(4)
𝑥→1
SAMPLE PROBLEMS | Limit of the Quotient of
=( lim 2𝑥) ( lim 4) Functions
𝑥→1 𝑥→1
=2( lim 𝑥) ( lim 4)
𝑥→1 𝑥→1 𝑥
1.​ lim ( 3 )
=(2(1))(40 𝑥→1
=(2)(4) lim 𝑥
𝑥→2
=8 = lim 3
𝑥→2

4
 1
= 3
Limit of a Radical
1
= 3
The limit of a radical function involves
evaluating the behavior of a function containing
𝑥−2
2.​ lim ( 6
) square roots or other roots as the input
𝑥→2 approaches a specific value.
lim 𝑥− lim 2
𝑥→2 𝑥→2
= lim 6 If lim 𝑓(𝑥) = 𝐿, then
𝑥→2
𝑥→𝑎
2−2
= 6 lim
𝑛
𝑓(𝑥) = 𝑛 lim 𝑓(𝑥)
4 𝑥→𝑎 𝑥→𝑎
= 6 𝑛
= 𝐿

3.​ lim ( 2 )
𝑥+3
SAMPLE PROBLEMS | Limit of a Radical
𝑥 → 3 1−𝑥
lim 𝑥+ lim 3 1.​ lim 2𝑥
𝑥→3 𝑥→3
= 2
𝑥→2
lim 1−( lim )
𝑥→3 𝑥→3 = lim 2𝑥
3+3 𝑥→2
= 2
1−(3) = 2 lim 𝑥
6 𝑥→2
=- 8
= 2(2)
= 4
2
4.​ lim ( ) =2
𝑥
𝑥→
1
2
lim 2 2.​ lim 18 − 𝑥
𝑥→
1
𝑥→9
= lim 𝑥
2

𝑥→
1
2 = lim 18 − lim 𝑥
2 𝑥→9 𝑥→9
= 1
(2) = 18 − 9
=4 = 9
=3
2
𝑥 −2𝑥+6
5.​ lim ( 2 ) 3.​ lim 𝑥 − 8
3

𝑥→1 𝑥 +4 𝑥 → 16
2
( lim 𝑥) − lim 2𝑥+ lim 6 = 3 lim 𝑥 − lim 8
𝑥→1 𝑥→1 𝑥→1
= 2
𝑥 → 16 𝑥 → 16
( lim 𝑥) + lim 4 3
𝑥→1 𝑥→1 = 16 − 8
2 3
( lim 𝑥) −2 lim 𝑥+ lim 6 = 8
𝑥→1 𝑥→1 𝑥→1
= 2
=2
( lim 𝑥) + lim 4
𝑥→1 𝑥→1
2 2
(1) −2(1)+6 𝑥 +4𝑥+4
= 2
4.​ lim 2
(1) +4 𝑥→0 𝑥 +1
1−2+6
= 1+4 2
( lim 𝑥) + lim 4𝑥+ lim 4
𝑥→0 𝑥→0 𝑥→0
5 =
= 5
2
( lim 𝑥) + lim 1
𝑥→0 𝑥→0
=1
5
 2
𝑥 −4
2
( lim 𝑥) +4 lim 𝑥+ lim 4 = lim 𝑥−2
𝑥→0 𝑥→0 𝑥→0 𝑥→2
= 2
( lim 𝑥) + lim 1 (𝑥+2)(𝑥−2)
𝑥→0 𝑥→0 = lim 𝑥−2
𝑥→2
2
(0) +4(0)+4
= 2
= lim x+2
(0) +1 𝑥→2

=
4 =2+2
1
=4
= 4
=2 𝑥+1−1
2.​ lim 𝑥
𝑥→0
2
5.​ lim
𝑥 +12−3 Directly substituting x will result to:
2
𝑥→2 𝑥 −1 0+1−1
= 0
2
( lim 𝑥) + lim 12− lim 3
𝑥→2 𝑥→2 𝑥→2
= 0, indeterminate
= 2
Solve by conjugation
( lim 𝑥) − lim 1
𝑥→2 𝑥→2
𝑥+1−1
2 = lim 𝑥
(2) +12−3
= 𝑥→0
2
(2) −1
𝑥+1−1 𝑥+1+1
4+12−3 = lim ⋅
= 4−1 𝑥→0
𝑥 𝑥+1+1

(𝑥+1)−1
=
16−3 = lim
3 𝑥( 𝑥+1+1)
𝑥→0
4−3
= 3 = lim
𝑥
𝑥( 𝑥+1+1)
1 𝑥→0
= 3 1
= lim
( 𝑥+1+1
𝑥→0
Evaluation of Limits that leads to =
1
( 0+1+1)
Indeterminate Forms 1
= 1+1
For certain rational functions, directly 1
=
evaluating their limits as x approaches a specific 2
value may not be simple. This happens when
direct substitution results in indeterminate forms 𝑥 −1
3

0 +∞ +∞ −∞ −∞
3.​ lim 2
like 0 , +∞
, −∞
, +∞
, −∞
, ∞ − ∞, 0 · (+ ∞), 𝑥 → 1 𝑥 −1

𝑎𝑛𝑑 0 · (− ∞). Directly substituting x will result to:

3
(1) −1
= 2
(1) −1
SAMPLE PROBLEMS | Indeterminate Forms = 0, indeterminate
Solve by factoring
2 3
𝑥 −4 𝑥 −1
1.​ lim 𝑥−2 = lim 2
𝑥→2 𝑥→1 𝑥 −1

Directly substituting x will result to: 2

(𝑥−1)(𝑥 +𝑥+1)
2 = lim (𝑥−1)(𝑥+1)
(2) −4 𝑥→1
= 2−2 2
𝑥 +𝑥+1
=
0 = lim 𝑥+1
0 𝑥→1
= 0, indeterminate 2
(1) +1+1
Solve by factoring = 1+1

6
 =2
3 Left hand side limits

2
lim 𝑓(𝑥) = 𝐿
−
𝑥 +𝑥−6 𝑥→𝑎
4.​ lim
𝑥 → 2 𝑥 −2 Provided we can make f(x) as close to L as
we want for all x sufficiently close to a with
Directly substituting x will result to: x<a without actually letting x be a.
2
(2) +2−6
=
2 −2 Right hand side limits
=0, indeterminate
Solve by factoring lim 𝑓(𝑥) = 𝐿
+
2
𝑥 +𝑥−6 𝑥→𝑎
= lim Provided we can make f(x) as close to L as
𝑥→2 𝑥 −2
we want for all x sufficiently close to a with
(𝑥+3)(𝑥−2)
= lim x>a without actually letting x be a.
𝑥→2 𝑥 −2

= lim (x+3) Existence of Limits

𝑥→2
=2+3 ​ Given a function f(x) if,
=5 lim 𝑓(𝑥) = lim 𝑓(𝑥) = 𝐿
+ −
𝑥→𝑎 𝑥→𝑎
2
𝑥 −16
5.​ lim then the normal limit will exist and
𝑥−2
𝑥→4
lim 𝑓(𝑥 = 𝐿)
Directly substituting x will result to: 𝑥→𝑎
2
(4) −16
=
4−2
SAMPLE PROBLEMS | Existence of Limits
=0 ,indeterminate
Solve by Conjugating and factoring
{ } find the limit of
2
𝑥, 𝑥<1
2
𝑥 −16 1.​ 𝑓(𝑥) =
= lim 1, 𝑥≥1
𝑥−2
𝑥→4 𝑓(𝑥) as 𝑥. approaches 1
(𝑥+4)(𝑥−4) 𝑥+2
= lim ⋅
𝑥→4
𝑥−2 𝑥+2 To find the left hand side limit
2
(𝑥+4)(𝑥−4)( 𝑥+2) lim 𝑓(𝑥) = lim (𝑥)
= lim 𝑥−4 𝑥→1
−
𝑥→1
−

𝑥→4
2
=1
= lim (𝑥 + 4)( 𝑥 + 2)
𝑥→4
= 1
To find the right hand side limit
=(4 + 4)( 4 + 2) lim 𝑓(𝑥) = lim 1
=(8)(4) 𝑥→1
+
𝑥→1
+

=32 = 1
Since, lim 𝑓(𝑥) = lim 𝑓(𝑥) then
+ −
𝑥→1 𝑥→1
One Sided Limits lim 𝑓(𝑥) = lim 𝑓(𝑥) = lim 𝑓(𝑥)
+ −
𝑥→1 𝑥→1 𝑥→1
One-sided limits describe the Thus the limit exists as x approaches 1.
behavior of a function as the input
approaches a specific point from only one 2.​ . 𝑓(𝑥) = { 2𝑥−1,
2𝑥+1,
𝑥 <2
𝑥≥2 } find the limit
direction—either from the left or the right. of 𝑓(𝑥) as 𝑥. approaches 2
They are essential for analyzing
discontinuities, continuity, and endpoints in To find the left hand side limit
functions.
lim 𝑓(𝑥) = lim (2𝑥 − 1)
− −
𝑥→2 𝑥→2

7
 = (2(2) − 1) X-VALUES Y-VALUES
=4-1
=3 1.0001 1.00020001
To find the right hand side limit
1.001 1.002001
lim 𝑓(𝑥) = lim (2𝑥 + 1)
+ +
𝑥→2 𝑥→2
1.01 1.0201
= (2(2) + 1)
= 4 +1 1.1 1.21
=5
Since, lim 𝑓(𝑥) ≠ lim 𝑓(𝑥) then ≈ 1
+ −
𝑥→2 𝑥→2
lim 𝑓(𝑥) ≠ lim 𝑓(𝑥) ≠ lim 𝑓(𝑥) Since, lim 𝑓(𝑥) = lim 𝑓(𝑥) then
+ −
𝑥→2 𝑥→2 𝑥→2 + −
𝑥→1 𝑥→1
Thus the limit does not exist as x approaches 2.
lim 𝑓(𝑥) = lim 𝑓(𝑥) = lim 𝑓(𝑥)
+ −
𝑥→1 𝑥→1 𝑥→1

{ } find the limit of

2
𝑥, 𝑥 <4 Thus the limit does not exist as x approaches 1.
3.​ 𝑓(𝑥) = 2𝑥, 𝑥≥4

𝑓(𝑥) as 𝑥. approaches 4 1
5.​ lim 𝑓(𝑥) = 𝑥−3 as find the limit of
𝑥→3
To find the left hand side limit 𝑓(𝑥) as 𝑥 approaches 3
2
lim 𝑓(𝑥) = lim (𝑥)
𝑥→4
−
𝑥→4
− To find the left limit
2
=4
X-VALUES Y-VALUES
= 16
To find the right hand side limit 2.7 -3.33
lim 𝑓(𝑥) = lim 2𝑥
+ +
𝑥→4 𝑥→4 2.8 -5
= 2(4)
2.9 -10
=8
Since, lim 𝑓(𝑥) ≠ lim 𝑓(𝑥) then 2.99 -100
+ −
𝑥→4 𝑥→4
lim 𝑓(𝑥) ≠ lim 𝑓(𝑥) ≠ lim 𝑓(𝑥) ≈ -∞
+ −
𝑥→4 𝑥→4 𝑥→4
Thus the limit does not exist as x approaches 4.
To find the right limit
2
4.​ lim 𝑓(𝑥) = 𝑥 as find the limit of
𝑥→1 X-VALUES Y-VALUES
𝑓(𝑥) as 𝑥 approaches 1 3.3 3

To find the left limit 3.2 5

3.1 10
X-VALUES Y-VALUES
3.01 100
0.0009 0.00000081
≈ +∞
0.009 0.000081

0.09 0.0081 Since, lim 𝑓(𝑥) ≠ lim 𝑓(𝑥) then

+ −
𝑥→4 𝑥→4

0.9 0.81 lim 𝑓(𝑥) ≠ lim 𝑓(𝑥) ≠ lim 𝑓(𝑥)

+ −
𝑥→4 𝑥→4 𝑥→4
≈ 1 Thus the limit does not exist as x approaches 3.

To find the right limit

8
 Graph:
Infinite Limits
We Say,
lim 𝑓(𝑥) =+ ∞
−
𝑥→𝑎
If we can make f(x) arbitrarily large and
positive for all x sufficiently close to x = a,
from both sides, without actually letting x = a.

We Say,
lim 𝑓(𝑥) = − ∞
−
𝑥→𝑎
If we can make f(x) arbitrarily large and
negative for all x sufficiently close to x = a, 1
2.​ lim 𝑓(𝑥) =
from both sides, without actually letting x = a. 𝑥→0
𝑥

SAMPLE PROBLEMS | Using Tabular and Find The Left Side Limit
Graphical Approach
X-VALUES Y-VALUES
1
1.​ lim 𝑓(𝑥) =
𝑥→0
2
𝑥 -0.5 DNE

-0.75 DNE
Find The Left Side Limit
-0.99 DNE
X-VALUES Y-VALUES
DNE
-0.5 4
Find The Right Side Limit
-0.75 1.79

-0.99 1.02
X-VALUES Y-VALUES
+∞
0.5 1.41
Find The Right Side Limit
0.75 1.155

X-VALUES Y-VALUES 0.99 1.005

0.5 4 +∞

0.75 1.79
As 𝑥 approaches 0 from both sides, only
0.99 1.02 the right hand limit increases towards +∞.
Thus, the limit does not exist as x
+∞ approaches to 0.

As x approaches 0 from both sides 𝑓(x) Graph

increases towards +∞, confirming the infinite
limit.

9
 1
4.​ lim 𝑓(𝑥) = 3
𝑥→0 𝑥

Find The Left Side Limit

X-VALUES Y-VALUES

-0.1 -1000

-0.01 -1000000

-0.001 -1000000000
3
3.​ lim 𝑓(𝑥) = 2
𝑥→0 𝑥 -∞

Find The Left Side Limit Find The Right Side Limit

X-VALUES Y-VALUES X-VALUES Y-VALUES

-0.1 300 0.1 1000

-0.01 30000 0.01 1000000

-0.001 3000000 0.001 1000000000

+∞ +∞

Find The Right Side Limit As 𝑥 approaches 0 from both sides, only
the right hand limit increases towards +∞.
X-VALUES Y-VALUES Thus, the limit does not exist as x
approaches to 0.
0.1 300

0.01 30000 Graph

0.001 3000000

+∞

As x approaches 0 from both sides 𝑓(x)

increases towards +∞, confirming the infinite
limit.

Graph

1
5.​ lim 𝑓(𝑥) = 2
𝑥→2 (𝑥−2)

Find The Left Side Limit

X-VALUES Y-VALUES

1.5 4

1.9 100

10
 1
1.99 10000 lim 𝑟 = + ∞
+ (𝑥−𝑎)
𝑥→𝑎
-∞

SAMPLE PROBLEMS | Using Theorems

Find The Right Side Limit

1
X-VALUES Y-VALUES 1.​ lim 𝑥−1
+
𝑥→1
2.01 4 lim 1
+
𝑥→1
=
lim + 𝑥 − lim 1
2.1 100 𝑥→1 𝑥→1
+

1
2.5 10000 = 1−1
1
+∞
= 0
, undefined

As x approaches 2 from both sides 𝑓(x) However, from the theorem, if 𝑟 is odd,
1
increases towards +∞, confirming the infinite lim 𝑟 = + ∞. Since r =1 which is
𝑥→𝑎
+ (𝑥−𝑎)
limit.
1
​ an odd number, lim 𝑥−1
= + ∞
+
𝑥→1
Graph

Graph

Using Theorems
1
If r is even 2.​ lim
+ (𝑥−2)
𝑥→2

1
lim 𝑟 = + ∞ lim 1
+
(𝑥−𝑎) 𝑥→2
𝑥→𝑎 =
lim 𝑥−2
1 𝑥→2
+

lim 𝑟 = + ∞
+ (𝑥−𝑎)
𝑥→𝑎 lim 1
+
𝑥→2
=
lim 𝑥− lim 2
If r is odd 𝑥→2
+
𝑥→2
+

1
1 =
lim 𝑟 = − ∞ 2−2
𝑥→𝑎
− (𝑥−𝑎)
1
1 =
lim 𝑟 = + ∞ 0
𝑥→𝑎
+ (𝑥−𝑎)
1
1 = , undefined.
lim 𝑟 = − ∞ 0
− (𝑥−𝑎)
𝑥→𝑎

11
However, from the theorem, if 𝑟 is even,
1
lim = + ∞. Since r = 2, which is
However, from the theorem, if 𝑟 is odd,
𝑟
+ (𝑥−𝑎)
𝑥→𝑎
1
an even number, lim
1
= + ∞. lim 𝑟 = − ∞. Since r = 3, which is
+ (𝑥−4) 𝑥→𝑎
− (𝑥−𝑎)
𝑥→2
1
an odd number, lim 3 = − ∞.
𝑥→2
− (𝑥−2)
Graph

Graph

1
4.​ lim 3
− (𝑥−3)
𝑥→3

lim 1
−
𝑥→3
= 3
lim (𝑥−3)
−
𝑥→3

lim 1
−
𝑥→3
=
3 lim 𝑥− lim 3
− −
𝑥→3 𝑥→3

1
3.​ lim 3 =
1
𝑥→2
− (𝑥−2 ) 3
3−3

lim 1 1
−
= 3
𝑥→2 0
= 3
lim (𝑥−2)
𝑥→2
− 1
= 0
, undefined.

lim 1
𝑥→2
− However, from the theorem, if 𝑟 is even,
= 1
3
lim = − ∞. Since r = 3, which is
( lim 𝑥 − lim 2
𝑥→2
−
𝑥→2
− ) 𝑥→𝑎
−
𝑟
(𝑥−𝑎)

an odd number, lim 3

1
= − ∞.
− (𝑥−5)
𝑥→3
1
= 3
(2−2)
Graph
1
= 3
(0)

1
= 0
, undefined.

1
5.​ lim 3
𝑥→3
+ (𝑥−3 )

lim 1
+
𝑥→3
= 3
lim (𝑥−3)
+
𝑥→3

12
 lim 1
+
𝑥→3
= 3

(lim 𝑥 − lim 3
𝑥→3
+
𝑥→3
+ ) If lim 𝑓(𝑥) =+ ∞, and lim ℎ(𝑥) =− ∞,
𝑥→𝑎 𝑥→𝑎
1 then,
= 3
(3−3)
lim [𝑓(𝑥) − ℎ(𝑥)] =− ∞
1
= 3 𝑥→𝑎
(0)

1
= 0
, undefined.
SAMPLE PROBLEMS | Sum
However, from the theorem, if 𝑟 is odd,
1.​ lim ⎡⎢ ⎤
1 1
lim 𝑟 = + ∞. Since r = 3, which is 2 + 2⎥
(𝑥−𝑎) 𝑥→3 ⎣ (𝑥−3) ⎦
+
𝑥→𝑎
1
an odd number, lim 3 = + ∞. ⎡ 1 ⎤ ⎡ ⎤
𝑥→3
+ (𝑥−3) = ⎢ lim ⎥ + ⎢ lim 2⎥
⎢𝑥 → 3 (𝑥−3) ⎥ ⎢𝑥 → 3 ⎥
2
⎣ ⎦ ⎣ ⎦
Graph =∞ + 2
=∞

2.​ lim ⎡⎢ ⎤
1 4
2 + 2 ⎥
𝑥→5 ⎣ (𝑥−5) (𝑥−5) ⎦
⎡ 1 ⎤ ⎡ 1 ⎤
= ⎢ lim ⎥+ 4⎢ lim ⎥
⎢𝑥 → 5 (𝑥−5)2 ⎥ ⎢𝑥 → 5 (𝑥−5)
2
⎥
⎣ ⎦ ⎣ ⎦
= ∞ + 4(∞)
=∞

3.​ lim ⎡⎢ ⎤
1 1
2 + (𝑥−1) ⎥
𝑥 → 1 ⎣ (𝑥−1) ⎦
Properties of Infinite Limits (Sum)
⎡ 1 ⎤ ⎡ 1 ⎤
= ⎢ lim ⎥ + ⎢ lim ⎥
⎢𝑥 → 1 (𝑥−1)2 ⎥ ⎢𝑥 → 1 (𝑥−1) ⎥
⎣ ⎦ ⎣ ⎦
If lim 𝑓(𝑥) =+ ∞, and lim 𝑔(𝑥) = 𝐿, then,
𝑥→𝑎 𝑥→𝑎
= (∞) + (∞)
=∞
lim [𝑓(𝑥) + 𝑔(𝑥)] =+ ∞
𝑥→𝑎
4.​ lim ⎡⎢− ⎤
1 1
2 − 2 ⎥
𝑥→3 ⎣ (𝑥−3) (𝑥−3) ⎦
If lim 𝑓(𝑥) =− ∞, and lim 𝑔(𝑥) = 𝐿, then, =
𝑥→𝑎 𝑥→𝑎
⎡ ⎤ ⎡ ⎤
⎢ lim − 1 2 ⎥ + ⎢ lim − 1
⎥
lim [𝑓(𝑥) + 𝑔(𝑥)] =− ∞ ⎢𝑥 → 3 (𝑥−3) ⎥ ⎢ (𝑥−3)
2
⎥
𝑥→𝑎
⎣ ⎦ ⎣𝑥 → 3 ⎦
=− ∞ − ∞
= − ∞

If lim 𝑓(𝑥) =+ ∞, and lim ℎ(𝑥) =− ∞ ,

5.​ lim ⎡⎢− − 2𝑥⎤⎥
𝑥→𝑎 𝑥→𝑎
1
6
then. 𝑥→3 ⎣ (𝑥−3) ⎦

lim [𝑓(𝑥) − ℎ(𝑥)] =+ ∞ ⎡ 1 ⎤ ⎡ ⎤

= ⎢ lim ⎥− 2⎢ lim 𝑥⎥
𝑥→𝑎 ⎢𝑥 → 3 (𝑥−3)
6
⎥ ⎢𝑥 → 3 ⎥
⎣ ⎦ ⎣ ⎦
13
 = − ∞ − 2(3) =+ ∞
=− ∞

Properties of Infinite Limits 2.​ lim ⎡⎢ 3

1
·
𝑥−4
𝑥−3
⎤
⎥
𝑥→2 ⎣ ⎦
+ 𝑥−2
(Product)
= lim ⎡⎢
𝑥→2 ⎣
+
3
𝑥−2
1 ⎤ · lim
⎥
⎦ 𝑥 → 2+
( )
𝑥−4
𝑥−3

If lim 𝑓(𝑥) =+ ∞, and lim 𝑔(𝑥) = 𝐿 > 0,

𝑥→𝑎 𝑥→𝑎 lim 𝑥− lim 4
+ +
then. 𝑥→2 𝑥→2
=+ ∞ · lim 𝑥− lim 3
+ +
𝑥→2 𝑥→2
lim [𝑓(𝑥)𝑔(𝑥)] =+ ∞
𝑥→𝑎 2−4
=+ ∞ · 2−3
If lim 𝑓(𝑥) =+ ∞, and lim 𝑔(𝑥) = 𝐿 < 0,
𝑥→𝑎 𝑥→𝑎
=+ ∞ ·2
then,

lim [𝑓(𝑥)𝑔(𝑥)] =− ∞ =+ ∞
𝑥→𝑎
2
3.​ lim ⎡⎢ ⎤
1
If lim 𝑓(𝑥) =+ ∞, and lim ℎ(𝑥) =+ ∞, 2 ⎥
𝑥→𝑎 𝑥→𝑎 𝑥 → −2 ⎣ (𝑥+2) ⎦
then,

= lim ⎡⎢ ⎤ lim ⎡ 1 ⎤
1
2 ⎥· ⎢ 2 ⎥
lim [𝑓(𝑥)ℎ(𝑥)] =+ ∞ 𝑥 → −2 ⎣ (𝑥+2) ⎦ 𝑥 → −2 ⎣ (𝑥+2) ⎦
𝑥→𝑎

If lim 𝑓(𝑥) =+ ∞, and lim ℎ(𝑥) =− ∞, = + ∞ · (+ ∞)

𝑥→𝑎 𝑥→𝑎
then, =+ ∞
lim [𝑓(𝑥)ℎ(𝑥)] =− ∞
lim ⎡⎢ ⎤
1 1
𝑥→𝑎 4.​ 3 · 2 ⎥
𝑥−2
𝑥 → −2 ⎣ (𝑥−2) ⎦
If lim 𝑓(𝑥) =− ∞, and lim ℎ(𝑥) =− ∞,

( )
𝑥→𝑎 𝑥→𝑎
1 1
then, = lim ⎡⎢ 3
⎤ · lim
⎥ 2
𝑥 → −2 ⎣ ⎦ 𝑥 → −2
𝑥−2 (𝑥−2)

lim [𝑓(𝑥)ℎ(𝑥)] =+ ∞
𝑥→𝑎 = − ∞ · (+ ∞)

=− ∞
SAMPLE PROBLEMS | Product
lim ⎡⎢ ⎤
1 1
5.​ 3 · 5 ⎥
𝑥+5
𝑥 → −5 ⎣ (𝑥+5) ⎦
−
2
1.​ lim ⎡⎢ ⎤
1 𝑥 +4
2 · 𝑥 ⎥
𝑥→2 ⎣ (𝑥−2) ⎦

= lim ⎢⎡
1 ⎤
2 ⎥ · lim ( )
2
𝑥 +4
𝑥
= lim ⎡⎢
𝑥 → −5 ⎣
−
1
(𝑥+5)
⎤
3 ⎥· lim
⎦ 𝑥 → −5−
( ) 5
1
𝑥+5
𝑥→2 ⎣ (𝑥−2) ⎦ 𝑥→2
2
lim 𝑥 + lim 4 = − ∞ · (− ∞)
𝑥→2 𝑥→2
=+ ∞ · lim 𝑥
𝑥→2 =+ ∞
2
2 +4
=+ ∞ · 2
8
=+ ∞ · 2
14
 lim 1
𝑥→0
=
Properties of Infinite Limits lim
−1
4
𝑥→0 𝑥
(Quotient) 1
= −∞

If lim 𝑓(𝑥) =+ ∞, and lim 𝑔(𝑥) = 𝐿, then,

=0
𝑥→𝑎 𝑥→𝑎 3−𝑥
5.​ lim 1

𝑔(𝑥) 𝑥→2 (𝑥−2)

2

lim 𝑓(𝑥)
= 0. lim 3+ lim 𝑥
𝑥→𝑎 𝑥→2 𝑥→2
= 1
lim 2
𝑥 → 2 (𝑥−2)

If lim ℎ(𝑥) =− ∞, and lim 𝑔(𝑥) = 𝐿, then 3+2

𝑥→𝑎 𝑥→𝑎
= +∞
5
𝑔(𝑥) = +∞
lim ℎ(𝑥)
= 0.
𝑥→𝑎 =0

SAMPLE PROBLEMS | Quotient

1.​ lim
3𝑥+3
1
Limits at Infinity
𝑥→0 𝑥
2

lim 3𝑥 + lim 3
𝑥→0 𝑥→0 Limits at infinity describe the behavior of
= 1 a function as the input variable (x) grows
lim 2
𝑥→0 𝑥
arbitrarily large (x→∞) or becomes arbitrarily
3(0) + 3
= +∞
small (x→−∞).
2
= +∞
Let k be any real number and r be any positive
rational number, then
=0
2 𝑘
𝑥 lim = 0
2.​ lim 1 𝑥 → +∞
𝑟
𝑥
𝑥→0 𝑥
4

2
lim 𝑥 𝑘 𝑟
𝑥→0 lim = 0, provided 𝑥 is defined
=
𝑟
𝑥 → −∞ 𝑥
1
lim 4
𝑥→0 𝑥 when using 𝑥 < 0.
0
= +∞
=0 SAMPLE PROBLEMS | Limits at Infinity
143 2
3.​ lim −1
1.​ lim (𝑥 − 3)
𝑥→0 2 𝑥→∞
𝑥
lim 143 Directly substituting x will result to:
𝑥→0 2
= −1
=(( lim 𝑥) − lim 3)
lim 2 𝑥→∞ 𝑥→∞
𝑥→0 𝑥
=∞-3
143
= −∞ =∞
=0 1
1 2.​ lim 𝑥
4.​ lim −1 𝑥→∞
𝑥→0 𝑥
4
Directly substituting x will result to:

15
 lim 1 2
lim 6𝑥 − lim 𝑥
𝑥→∞ 𝑥→∞ 𝑥→∞
= = 4
lim 𝑥 lim 3 + lim
𝑥→∞ 𝑥
𝑥→∞ 𝑥→∞

= ∞1 = 6(∞) −0
3−0
=0 ∞
= 3

3.​ lim
2
2𝑥 −2 =∞
2
𝑥→∞ 𝑥

Directly substituting x will result to: 2

5𝑥 − 2𝑥 + 10
2 5.​ lim 2
( lim 2𝑥) − lim 2 𝑥→∞ 2𝑥 + 3𝑥−6
𝑥→∞ 𝑥→∞
= 2 Directly substituting x will result to:
( lim 𝑥) 2
𝑥→∞ ( lim 5𝑥) − lim 2𝑥 + lim 10
𝑥→∞ 𝑥→∞ 𝑥→∞
=∞ = 2
∞ ( lim 5𝑥) + lim 3𝑥− lim 6
𝑥→∞ 𝑥→∞ 𝑥→∞
= indeterminate
∞−∞+10
But we can simplify the function by multiplying both = ∞+∞−6
numerator and denominator
2
= indeterminate
2𝑥 − 2
= lim 2
But we can simplify the function by multiplying both
𝑥→∞ 𝑥
numerator and denominator
1 2
2 5𝑥 − 2𝑥 + 10
= lim (
2𝑥 − 2
⋅
2
𝑥
) lim 2
𝑥
2 1 𝑥→∞ 2𝑥 + 3𝑥−6
𝑥→∞ 2
𝑥 1
2 2
2− 5𝑥 − 2𝑥 + 10 2
2
= lim ( ⋅ 𝑥
)
= lim ( 1
𝑥
) 𝑥→∞
2
2𝑥 + 3𝑥−6
1
𝑥
2
𝑥→∞
2 10
1 5− 𝑥
+ 2
lim 2 − lim
𝑥→∞ 𝑥→∞
2
𝑥 = lim ( 𝑥
)
=( lim 1
) 𝑥→∞ 2 +
3
𝑥
−
𝑥
6
2

𝑥→∞ 2 10
lim 5 − lim 𝑥
+ lim 2
2
= 1 =
𝑥→∞ 𝑥→∞ 𝑥→∞ 𝑥

3 6
lim 2 + lim − lim
=2 𝑥→∞ 𝑥→∞
𝑥
𝑥→∞ 𝑥
2

5−0+0
2 = 2+0−0
6𝑥 − 2
4.​ lim 5
𝑥 → ∞ 3𝑥 + 4 = 2
Directly substituting x will result to:
2
( lim 6𝑥) − lim 2
𝑥→∞ 𝑥→∞
Properties of Infinite Limits at
=
( lim 3𝑥) + lim 4 Infinity
𝑥→∞ 𝑥→∞
∞−2
= ∞+4
If the values of f(x) become extremely
large when x approaches:
= indeterminate
But we can simplify the function by multiplying both positive infinity (+ ∞)
numerator and denominator lim 𝑓(𝑥) =+ ∞
6𝑥 − 2
2 𝑋 → +∞
= lim
𝑥→∞ 3𝑥 + 4
negative infinity (− ∞)
1
6𝑥 − 2
2
lim 𝑓(𝑥) =+ ∞
= lim ( ⋅ 𝑥
1 ) 𝑋 → −∞
𝑥→∞ 3𝑥 + 4 𝑥

2
6𝑥 −
= lim ( 𝑥
4
) If the values of f(x) become extremely
𝑥→∞ 3 + 𝑥
small when x approaches:

16
 ●​ If 𝑏 > 0 𝑎𝑛𝑑 𝑏 ≠ 1, then the exponential
positive infinity (+ ∞) function with base b is the function f defined
lim 𝑓(𝑥) =− ∞ 𝑥
𝑋 → +∞ by 𝑓(𝑥) = 𝑏 .

negative infinity (− ∞) 𝑥
●​ If 𝑏 = 𝑒, 𝑓(𝑥) = 𝑒 is called the Natural
lim 𝑓(𝑥) =+ ∞ Exponential Function, where e is the Euler’s
𝑋 → −∞
number.

SAMPLE PROBLEMS | Properties of Infinite

Graph
Limits at Infinity

1.​ lim 4𝑥
𝑥→∞
= 4 lim 𝑥
𝑥→∞
= 4(∞)
=+∞

2. lim (2𝑥 + 2)
𝑥→∞
= 2 lim x + lim 2
𝑥→∞ 𝑥→∞
Properties of Exponential Functions
= 2(∞) + 2
=+∞ ●​ The graph of the function intersects the
y-axis at the point (0,1).
3. lim 4𝑥 ●​ The x-axis is the horizontal asymptote of
𝑥 → −∞
the graph of the function.
= 4 lim 𝑥
𝑥 → −∞
●​ If b > 1, the function values increase as
= 4(− ∞) x increases.
●​ If 0 < b < 1, the function values
=− ∞
decrease as x increases.
4. lim (4𝑥 − 2)
𝑥 → −∞
= 4 lim x - lim 2
𝑥 → −∞ 𝑥 → −∞ 𝑥 𝑐
lim 𝑒 = 𝑒
= 2(− ∞) - 2 𝑥→𝑐
=− ∞
𝑥 𝑐
lim 𝑎 = 𝑎
3 𝑥→𝑐
5. lim 𝑥
𝑥 → −∞
3
=( lim 𝑥)
𝑥 → −∞
SAMPLE PROBLEMS | Limits of Exponential
3
= (− ∞) Functions
=− ∞

Limits of Exponential Functions 1.​ lim 𝑒

𝑥

𝑥→2

An exponential function is expressed as = lim 𝑒

𝑥
𝑥 𝑥→2
𝑓(𝑥) = 𝑎 , where the variable x appears as an
2
exponent. The shape of the exponential curve is =𝑒
determined by the function itself and varies
based on the value of x 𝑥−1
2.​ lim 𝑒 + 2
𝑥→2

17
 𝑥−1 𝑚
= lim 𝑒 + lim 2 =4 lim
𝑒 −1
𝑥→2 𝑥→2 𝑚
𝑥→0
= 𝑒+2
= 4(1)
𝑥−2
=4
3.​ lim 3
𝑥→4
6𝑥
4−2 6𝑒 − 6
=3 2.​ lim 𝑥
2 𝑥→0
=3 6𝑥
6(𝑒 − 6)
=9 = lim 𝑥
𝑥→0
6𝑥
𝑒 −1 6
1𝑥 = 6 lim ( ⋅ )
4.​ lim 𝑥 6
𝑥→2
3 𝑥→0
6𝑥
12 𝑒 −1
= = 6(6) lim 6𝑥
3
𝑥→0
1
= 9
Let 6x = v
𝑣
1𝑥 𝑒 −1
5.​ lim 5
= 6(6) lim 𝑣
𝑥 → −3 𝑥→0
= (6)(6)(1)
−3
1
= −3
5
= 36
1
3
1
= 1 2𝑥
5
3 3.​ lim 2𝑥
𝑥→0 𝑒 −1
1
= 1
= lim
2𝑥
⋅
2
125 2𝑥 2
𝑥→0 𝑒 −1
= 125 1 2𝑥
= 2
lim 2𝑥
𝑥→0 𝑒 −1
Limits of Exponential Functions Let 2x= q
(Special Cases) ​ =
1
lim
𝑞
2 𝑞
𝑥→0 𝑒 −1

( )= 1
𝑥
𝑒 −1
lim 1
𝑥→0
𝑥 ​ = 2
(1)
1
​ = 2

​ ​ lim
𝑥→0
( )= 1
𝑥
𝑥
𝑒 −1 3𝑥
𝑒 −1
4.​ lim 2𝑥
𝑥→0 𝑒 −1
3𝑥
1 𝑒 −1 𝑥
= ( lim 2𝑥 ⋅ lim 1
) 𝑥
SAMPLE PROBLEMS | Special Cases 𝑥→0 𝑒 −1 𝑥→0
3𝑥
𝑥 2 𝑒 −1 3
= ( lim 2𝑥 ( ) ⋅ lim ( 3 ))
𝑥→0 𝑒 −1 2 𝑥→0
𝑥
4𝑥
𝑒 −1
1.​ lim 1 2𝑥
3𝑥
𝑒 −1
𝑥→0
𝑥 ​ = ( 2 lim 2𝑥 ⋅ 3 lim 3𝑥
)
4𝑥 𝑥→0 𝑒 −1 𝑥→0
𝑒 −1 4
= lim 𝑥
⋅ 4 Let 2x = c, Let 3x = d
𝑥→0 𝑑
1 𝑐 𝑒 −1
𝑒 −1
4𝑥 = ( 2 lim 𝑐 ⋅ 3 lim 𝑑
= 4 lim 4𝑥 𝑥→0 𝑒 −1 𝑥→0
𝑥→0 1
= (1) ⋅ 3(1)
Let 4x = m 2

18
 =
3 ●​ If b > 1, the function values increase as
2
x increases.
●​ If 0 < b < 1, the function values
2
5.​ lim
𝑥 decrease as x increases.
6𝑥
𝑥→0 𝑒 −1

𝑥 lim 𝑙𝑛𝑥 = 𝑙𝑛𝑐

= lim 6𝑥 ⋅ lim 𝑥 𝑥→𝑐
𝑥→0 𝑒 −1 𝑥→0
𝑥 6 lim 𝑙𝑜𝑔𝑏𝑥 = 𝑙𝑜𝑔𝑏𝑐
= lim 6𝑥 ( 6 ) ⋅ lim 𝑥
𝑥→0 𝑒 −1 𝑥→0 𝑥→𝑐
1 6𝑥
= 6
lim 6𝑥 ⋅ lim 𝑥 SAMPLE PROBLEMS | Logarithmic
𝑥→0 𝑒 −1 𝑥→0
Let 6x = o Functions
1 𝑜
= 6
lim 𝑜 ⋅ lim 𝑥
𝑥→0 𝑒 −1 𝑥→0 1.​ lim 𝑙𝑛(𝑥)
1 𝑥→1
= ⋅0
6 = 𝑙𝑛(1)
=0 =0

Limit Of Logarithmic Functions 2.​ lim 𝑙𝑜𝑔9(9𝑥 − 720)

2

𝑥→9
2
​ A logarithmic function is the inverse of = 𝑙𝑜𝑔9(9(9) − 720)
an exponential function, expressed as
= 𝑙𝑜𝑔9(729 − 720)
𝑓(𝑥) = 𝑙𝑜𝑔 𝑏𝑥 , where b is the base. It
= 𝑙𝑜𝑔9(9)
represents the exponent to which b must be
raised to produce x. =1
when:
3
3.​ lim 𝑙𝑛(𝑥 − 9) + 34𝑥
●​ 𝑏 = e, 𝑓 (𝑥) is called Natural Logarithmic 𝑥→9
Function 3
= 𝑙𝑛((9) − 9) + 34(9)
●​ 𝑙𝑜𝑔 𝑒𝑥 = 𝑙𝑛 𝑥
= 𝑙𝑛(729 − 9) + 306
●​ 𝑏 = 10, 𝑓 (𝑥) is called Common = 𝑙𝑛(720) + 306
Logarithmic Function
●​ 𝑏 = 2, 𝑓 (𝑥) is called Binary Logarithmic 𝑙𝑛(𝑥−1)+3𝑥+9
Function 4.​ lim 2𝑥−1
𝑥→2
𝑙𝑛(2−1)+3(2)+9
= 2(2)−1
Graph 𝑙𝑛(1)+6+9
= 3
15
= 3
=5

5.​ lim − 𝑙𝑛(𝑥 − 𝑒)

𝑥 → 2𝑒
=− 𝑙𝑛((2𝑒) − 𝑒)
=− 𝑙𝑛(𝑒)
Properties of Logarithmic Functions =− 1

●​ The graph of the function intersects the

x-axis at the point (1,0).
Limit of Trigonometric
●​ The y-axis is the vertical asymptote of Functions
the graph of the function.
19
 Trigonometric functions have distinct =
1
2
behaviors when evaluating limits, depending
on the value that the variable approaches.
These limits can be calculated using direct Limit of Trigonometric
substitution, special trigonometric identities. Functions (Special Cases)

lim
𝑥→𝑐
𝑠𝑖𝑛 𝑥 = 𝑠𝑖𝑛 𝑐 lim
𝑥→0
( )= 1 𝑠𝑖𝑛𝑥
𝑥

lim ( )= 1 𝑥
lim 𝑐𝑜𝑠 𝑥 = 𝑐𝑜𝑠 𝑐 𝑠𝑖𝑛𝑥
𝑥→𝑐 𝑥 → 0

lim ( )= 0 1−𝑐𝑜𝑠𝑥
lim 𝑡𝑎𝑛 𝑥 = 𝑡𝑎𝑛 𝑐 𝑥
𝑥→𝑐 𝑥 → 0

lim 𝑐𝑠𝑐 𝑥 = 𝑐𝑠𝑐 𝑐

𝑥→𝑐

lim 𝑠𝑒𝑐 𝑥 = 𝑠𝑒𝑐 𝑐 SAMPLE PROBLEMS | Special Cases

𝑥→𝑐
𝑠𝑖𝑛2𝑥
lim 𝑐𝑜𝑡 𝑥 = 𝑐𝑜𝑡 𝑐 1.​ lim 𝑥
𝑥→𝑐 𝑥→0
𝑠𝑖𝑛2𝑥 2
= lim ( 𝑥
⋅ 2
)
𝑥→0

SAMPLE PROBLEMS | Trigonometric = 2 lim

𝑠𝑖𝑛2𝑥
2𝑥
𝑥→0
Functions
𝑐𝑜𝑠(𝑥) Let 2x = w
1.​ lim 2 𝑠𝑖𝑛 𝑤
𝑥→ 2
π 𝑠𝑖𝑛 (𝑥) = 2 lim 𝑤
𝑥→0
π
𝑐𝑜𝑠( 2 )
= = 2 (1)
2 π
𝑠𝑖𝑛 ( 2 ) =2
=0
𝑠𝑖𝑛3𝑥
2.​ lim 4𝑥
𝑥→0
2.​ lim 𝑠𝑒𝑐(𝑥)
𝑠𝑖𝑛 3𝑥 3
𝑥→π = lim ( ⋅ )
4𝑥 3
𝑥→0
= 𝑠𝑒𝑐(π)
3 𝑠𝑖𝑛 3𝑥
= -1 = 4
lim 3𝑥
𝑥→0
Let 3x = y
2 2
3.​ lim 𝑐𝑜𝑠 (𝑥) − 𝑠𝑖𝑛 (𝑥) =
3
lim
𝑠𝑖𝑛 𝑦
𝑥→π 4 𝑦
𝑥→0
2 2 3
= lim 𝑐𝑜𝑠 (𝑥) − lim 𝑠𝑖𝑛 (𝑥) = 4
(1)
𝑥→π 𝑥→π 3
2 2 = 4
= 𝑐𝑜𝑠 (1) − 𝑠𝑖𝑛 (1)
=1 𝑠𝑖𝑛2𝑥
3.​ lim 1−𝑐𝑜𝑠 4𝑥
𝑥→0
−3𝑡𝑎𝑛(𝑥) 𝑠𝑖𝑛2𝑥 1
4.​ lim 𝑠𝑒𝑐(𝑥) = lim ⋅
1 1−𝑐𝑜𝑠 4𝑥
𝑥→0 𝑥→0
−3𝑡𝑎𝑛(0) 𝑠𝑖𝑛2𝑥 1 𝑥
= = ( lim 1
⋅ lim 1−𝑐𝑜𝑠 4𝑥
)(𝑥)
𝑠𝑒𝑐(0)
𝑥→0 𝑥→0
=0 𝑠𝑖𝑛2𝑥 2
= ( lim 𝑥
⋅ 2
) ( lim
𝑥→0 𝑥→0
𝑠𝑖𝑛 𝑥 𝑥 4
5.​ lim 𝑥+2 1−𝑐𝑜𝑠 4𝑥
⋅ 4
)
𝑥→0 𝑠𝑖𝑛2𝑥 1 4𝑥
𝑠𝑖𝑛 (0) = (2 lim 2𝑥
)( 4
( lim 1−𝑐𝑜𝑠 4𝑥
)
= (0)+2 𝑥→0 𝑥→0

20
Let 2x = a, Let 4x = b
𝑠𝑖𝑛 𝑎 1 𝑏
= (2 lim 𝑎
)( 4
( lim 1−𝑐𝑜𝑠 𝑏
)
𝑥→0 𝑥→0
1
=(2(1))( 4 (0))
=0

4.​ lim
𝑥→0
( 1−𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥 )
= lim
𝑥→0
( 1−𝑐𝑜𝑠𝑥
1
·
1
𝑠𝑖𝑛𝑥
·
𝑥
𝑥 )
= lim
𝑥→0
( 1−𝑐𝑜𝑠𝑥
𝑥
·
𝑥
𝑠𝑖𝑛𝑥 )
= lim
𝑥→0
( 1−𝑐𝑜𝑠𝑥
𝑥 )· lim
𝑥→0
( ) 𝑥
𝑠𝑖𝑛𝑥

=0·1

=0

𝑠𝑖𝑛2𝑥
5.​ lim 6𝑥
𝑥→0
𝑠𝑖𝑛2𝑥 2
= lim 6𝑥
⋅ 2
𝑥→0
2 𝑠𝑖𝑛2𝑥
= 6
lim 2𝑥
𝑥→0
Let 2x = g
2 𝑠𝑖𝑛𝑔
​ = 6
lim 𝑔
𝑥→0
2
= 6
(1)​
2
= 6

21
 CONTINUITY

In this chapter, you should learn and meet continuity helps model drug concentration in
the following objectives: the bloodstream over time, ensuring safe
dosage regimens. Environmental science
Represent the continuity of a function applies continuous functions to simulate
at a specific point. climate change, pollutant diffusion, and
ecosystem dynamics. Computer graphics
Identify whether a function is and animation depend on continuity for
continuous at a given number. smooth rendering of curves, motion, and
visual effects. Statistics employs
Demonstrate various types of continuous probability distributions to
discontinuities. analyze real-valued data, while finance
uses them to model asset prices and risk.
Explain the continuity of a function Biology describes population growth and
over an interval. chemical reactions as continuous processes
when changes occur incrementally. By
Assess if a function remains ensuring functions behave predictably
continuous within a given interval. without disruptions, continuity provides a
framework for reliable analysis, innovation,
Apply continuity concepts to solve and decision-making across science,
mathematical problems. industry, and technology.

Introduction
In mathematics, continuity is a
fundamental concept that describes a
function whose output changes smoothly
without sudden jumps, breaks, or holes as
its input varies. A function is continuous if
small changes in the input result in small,
predictable changes in the output. Continuity
is essential in calculus and analysis, forming
the basis for defining derivatives, integrals,
and modeling real-world phenomena where
gradual, unbroken behavior is required. Continuity in economics and finance is important in modeling
Continuity underpins precise modeling and economic cycle and markets analysis. Economists represent
supply and demand curves, production functions, and utility
problem-solving across disciplines. In functions respectively as continuous functions. Using continuity
engineering, continuous functions describe analysis, economists can analyze economic systems stability,
stress distribution in materials, fluid flow, and tendency of trends in the market, and develop some solutions
and policy decisions.
electrical signals, ensuring stable designs.
Physics relies on continuity to model
Continuity of a Function at a
motion, wave propagation, and
thermodynamic processes, where abrupt number
changes are physically implausible.
Economics uses continuous demand and A function 𝑓(𝑥) is continuous at a
supply curves to predict market equilibrium point 𝑥 = 𝑎 if its behavior near a a aligns
and optimize pricing strategies. In medicine, seamlessly with its value at a a. This

22
concept ensures there are no abrupt jumps, three conditions does not meet at 𝑎, the
breaks, or holes in the function’s graph at function 𝑓 is considered discontinuous at 𝑎
that point. Formally, a function 𝑓 is .
considered continuous at a number 𝑎 if and
only if all three of the following conditions Continuity can be visualized using
are met: the "pencil test". If you can draw the
function’s graph around 𝑥 = 𝑎 without lifting
f(a) exists; your pencil, the function is continuous at
lim f(x) exists; that point. However, if there is a gap or
𝑥→𝑎 break in the graph, the function is
f(a) = lim f(x) discontinuous at that point.
𝑥→𝑎
If all three conditions hold, the
function is continuous at 𝑎. If any of these

SAMPLE PROBLEMS | Determine If Continuous Using The 3 Conditions

Determine if 𝑓(𝑥) = 4𝑥 is continuous at 𝑥 = 2.

Solution

a.​ Evaluate 𝑓 at 𝑥 = 2.
𝑓(𝑥) = 4𝑥
𝑓(2) = 4(2)
𝑓(2) = 8
Hence, 𝑓(3) exists.
b.​ Evaluate lim 4𝑥.
𝑥→3
lim 4𝑥 = lim 4 · lim 𝑥
𝑥→2 𝑥→2 𝑥→2
= 4 · 2
= 8
Thus, lim 4𝑥 exists.
𝑥→2
c.​ 𝑓(2) = lim 4𝑥 = 8
𝑥→2
Since all the conditions are satisfied, 𝑓 is continuous at 𝑥 = 2.

𝑥
Determine if 𝑓(𝑥) = 𝑥+1
is continuous at 𝑥 = 0.
Solution

a.​ Evaluate 𝑓 at 𝑥 = 0.
𝑥
𝑓(𝑥) = 𝑥+1
0
𝑓(0) = 0+1
𝑓(0) = 0
Hence, 𝑓(0) exists.
𝑥
b.​ Evaluate lim 𝑥+1
.
𝑥→0
lim 𝑥
𝑥 𝑥→0
lim 𝑥+1
= lim 𝑥 + lim 1
𝑥→0 𝑥→0 𝑥→0

0
= 0+1

23
 0
= 1
= 0
𝑥
Thus, lim 𝑥+1
exists.
𝑥→0
𝑥
c.​ 𝑓(0) = lim 𝑥+1 = 0
𝑥→0
Since all the conditions are satisfied, 𝑓 is continuous at 𝑥 = 0.

Determine if 𝑓(𝑥) = 𝑥 is continuous at 𝑥 = 0.

Solution

a.​ Evaluate 𝑓 at 𝑥 = 0.
𝑓(𝑥) = 𝑥
𝑓(0) = 0
𝑓(0) = 0
Hence, 𝑓(0) exists.
b.​ Evaluate lim 𝑥.
𝑥→0

lim 𝑥 = lim 𝑥
𝑥→0 𝑥→0

= 0
= 0
Thus, lim 𝑥 exists.
𝑥→0

c.​ 𝑓(0) = lim 𝑥 = 0

𝑥→0
Since all the conditions are satisfied, 𝑓 is continuous at 𝑥 = 0.

2
Determine if 𝑓(𝑥) = 𝑥 + 1 is continuous at 𝑥 = 0.
Solution

a.​ Evaluate 𝑓 at 𝑥 = 0.
2
𝑓(𝑥) = 𝑥 + 1
2
𝑓(0) = 0 + 1
𝑓(0) = 1
Hence, 𝑓(0) exists.
b.​ Evaluate lim 𝑥.
𝑥→0
2 2
lim 𝑥 + 1 = ( lim 𝑥) + lim 1
𝑥→0 𝑥→0 𝑥→0
2
= ( 0) + 1
= 0
Thus, lim 𝑥 exists.
𝑥→0
2
c.​ 𝑓(0) = lim 𝑥 + 1 = 1
𝑥→0
Since all the conditions are satisfied, 𝑓 is continuous at 𝑥 = 0.

24
 2
Determine if 𝑓(𝑥) = 3𝑥 − 4𝑥 + 5 is continuous at 𝑥 = 1.
Solution

a.​ Evaluate 𝑓 at 𝑥 = 1.
2
𝑓(𝑥) = 3𝑥 − 4𝑥 + 5
2
𝑓(1) = 3(1) − 4(1) + 5
𝑓(1) = 3 − 4 + 5
𝑓(1) = 4
Hence, 𝑓(1) exists.
2
b.​ Evaluate lim 3𝑥 − 4𝑥 + 5.
𝑥→1
2 2
lim 3𝑥 − 4𝑥 + 5 = (3 lim 𝑥) − 4 lim 𝑥 + lim 5
𝑥→1 𝑥→1 𝑥→1 𝑥→1
2
= 3(1) − 4(1) + 5
= 3−4+5
=4
2
Thus, lim 3𝑥 − 4𝑥 + 5 exists.
𝑥→1
2
c.​ 𝑓(1) = lim 3𝑥 − 4𝑥 + 5 = 4
𝑥→1
Since all the conditions are satisfied, 𝑓 is continuous at 𝑥 = 1.

Types of Discontinuity
1.​ Removable Discontinuity occurs if the
Discontinuities in mathematical lim 𝑓(𝑥) exists for a function (i.e.,
𝑥→𝑎
functions represent critical points where a
function's behavior deviates from continuity, lim 𝑓(𝑥) = lim 𝑓(𝑥)) but is not equal
𝑥→𝑎 𝑥→𝑎
resulting in breaks or unpredictable
to 𝑓(𝑎). This means that the first or third
characteristics in its graphical
condition is not satisfied.
representation. This section shows the three
fundamental types of discontinuities
recognized in mathematical analysis:
removable discontinuities, jump
discontinuities, and essential (infinite)
discontinuities.

25
 2.​ Jump Discontinuity occurs when the
left-hand lim 𝑓(𝑥) and the right-hand
−
𝑥→𝑎
lim 𝑓(𝑥) exists but are not equal for a
+
𝑥→𝑎
function 𝑓 somehow doesn’t exist, the
limit. Then 𝑥 = 𝑎𝑛is referred to as a
“jump discontinuity” or “non-removable
discontinuity.” This means that the
second condition is not satisfied.

3.​ Essential or Infinite Discontinuity

occurs if one or both of the limits of the
lim 𝑓(𝑥) and lim 𝑓(𝑥) is ± ∞ have
− +
𝑥→𝑎 𝑥→𝑎
the same value. It is known as “essential
discontinuity”or “infinite discontinuity.” In
such a discontinuity, one of the left-hand
and right-hand bounds cannot exist.

SAMPLE PROBLEMS | Removable/Hole Discontinuity

2
𝑥 −16
Determine if 𝑓(𝑥) = 𝑥−4
is continuous at 𝑥 = 4.
Solution

a.​ Evaluate 𝑓 at 𝑥 = 4.
2
𝑥 −16
𝑓(𝑥) = 𝑥−4
2
4 −16
𝑓(4) = 4−4
16−16
𝑓(4) = 4−4
0
𝑓(4) = 0
, which is indeterminate
b.​ Since 𝑓(4) is not defined, the first condition is not satisfied. Therefore, it has a
removable discontinuity, indicating that 𝑓 is discontinuous at 𝑥 = 4.

26
Graph

2
𝑥 −9
Determine if 𝑓(𝑥) = 𝑥−3
is continuous at 𝑥 = 3.
Solution

a.​ Evaluate 𝑓 at 𝑥 = 3.
2
𝑥 −9
𝑓(𝑥) = 𝑥−3
9−9
𝑓(3) = 3−3
0
𝑓(3) = 0
, which is indeterminate
b.​ Since 𝑓(3) is not defined, the first condition is not satisfied. Therefore, it has a
removable discontinuity, indicating that 𝑓 is discontinuous at 𝑥 = 3.
Graph

27
 2
2𝑥 −8
Determine if 𝑓(𝑥) = 𝑥−2
is continuous at 𝑥 = 2.
Solution

a.​ Evaluate 𝑓 at 𝑥 = 2.
2
2𝑥 −8
𝑓(𝑥) = 𝑥−2
2
2(2) −8
𝑓(2) = 2−2
2(4)−8
𝑓(2) = 2−2
8−8
𝑓(2) = 2−2
0
𝑓(2) = 0
, which is indeterminate
b.​ Since 𝑓(2) is not defined, the first condition is not satisfied. Therefore, it has a
removable discontinuity, indicating that 𝑓 is discontinuous at 𝑥 = 2.
Graph

3
𝑥 −1
Determine if 𝑓(𝑥) = 𝑥−1
is continuous at 𝑥 = 1.
Solution

a.​ Evaluate 𝑓 at 𝑥 = 1.
3
𝑥 −1
𝑓(𝑥) = 𝑥−2
3
1 −1
𝑓(1) = 1−1
1−1
𝑓(1) = 1−1
0
𝑓(1) = 0
, which is indeterminate
b.​ Since 𝑓(1) is not defined, the first condition is not satisfied. Therefore, it has a
removable discontinuity, indicating that 𝑓 is discontinuous at 𝑥 = 1.

28
Graph

3𝑥−6
Determine if 𝑓(𝑥) = 𝑥−2
is continuous at 𝑥 = 2.
Solution

a.​ Evaluate 𝑓 at 𝑥 = 2.
3𝑥−6
𝑓(𝑥) = 𝑥−2
3(2)−6
𝑓(2) = 2−2
6−6
𝑓(2) = 1−1
0
𝑓(2) = 0
, which is indeterminate
b.​ Since 𝑓(1) is not defined, the first condition is not satisfied. Therefore, it has a
removable discontinuity, indicating that 𝑓 is discontinuous at 𝑥 = 1.
Graph

29
SAMPLE PROBLEMS | Jump Discontinuity

Determine if 𝑓(𝑥) = { 3𝑥−1 ,

2
𝑥 +2 ,
𝑥≤1
𝑥>1 } is continuous at 𝑥 = 1.
Solution

a.​ From the given, 𝑥 = 1 is part of the the domain 𝑥 ≤ 1. Thus,

𝑓(𝑥) = 3𝑥 − 1
𝑓(1) = 3(1) − 1
𝑓(1) = 2
Hence, 𝑓(1) exists.
b.​ In evaluating lim 𝑓(𝑥), consider the two-sided limits.
𝑥→1
lim 𝑓(𝑥) = lim 3𝑥 − 1
− −
𝑥→1 𝑥→1
= 3(1) − 1
=2
2
lim 𝑓(𝑥) = lim 𝑥 + 2
+ +
𝑥→1 𝑥→1
2
= (1) + 2
=3
Since lim 𝑓(𝑥) ≠ lim 𝑓(𝑥), lim 𝑓(𝑥) does not exist. Therefore,
− +
𝑥→1 𝑥→1 𝑥→1
f has a jump discontinuity at 𝑥 = 1.
Graph

Determine if 𝑓(𝑥) = { 𝑥+3 ,

2−𝑥 ,
𝑥<0
𝑥 ≥0 } is continuous at 𝑥 = 0.
Solution

a.​ From the given, 𝑥 = 0 is part of the the domain 𝑥 ≥ 0. Thus,

𝑓(𝑥) = 2 − 𝑥
𝑓(0) = 2 − 0
𝑓(0) = 2
Hence, 𝑓(0) exists.

30
 b.​ In evaluating lim 𝑓(𝑥), consider the two-sided limits.
𝑥→0

lim 𝑓(𝑥) = lim 𝑥+3

− −
𝑥→0 𝑥→0

= 0+3
= 3
lim 𝑓(𝑥) = lim 2 − 𝑥
+ +
𝑥→0 𝑥→0
=2−0
=2
Since lim 𝑓(𝑥) ≠ lim 𝑓(𝑥), lim 𝑓(𝑥) does not exist.
− +
𝑥→0 𝑥→0 𝑥→0
Therefore, f has a jump discontinuity at 𝑥 = 0.
Graph

Determine if 𝑓(𝑥) = { |𝑥|

4−𝑥 ,
2
, 𝑥 ≤−2
𝑥 >−2 } is continuous at 𝑥 =− 2.
Solution

a.​ From the given, 𝑥 =− 2 is part of the the domain 𝑥 ≤− 2. Thus,

𝑓(𝑥) = |𝑥|
𝑓(− 2) = |− 2|
𝑓(− 2) = 2
Hence, 𝑓(− 2) exists.
b.​ In evaluating lim 𝑓(𝑥), consider the two-sided limits.
𝑥 → −2
lim 𝑓(𝑥) = lim |𝑥|
− −
𝑥 → −2 𝑥 → −2
= |− 2|
=2
2
lim 𝑓(𝑥) = lim 4 − 𝑥
+ +
𝑥 → −2 𝑥 → −2
2
= 4 − (− 2)
=0

31
 Since lim 𝑓(𝑥) ≠ lim 𝑓(𝑥), lim 𝑓(𝑥) does not exist.
− +
𝑥 → −2 𝑥 → −2 𝑥 → −2
Therefore, f has a jump discontinuity at 𝑥 =− 2.
Graph

Determine if 𝑓(𝑥) = { 2𝑥+5,

10−𝑥, } is continuous at 𝑥 = 3.
𝑥 ≤3
𝑥 >3
Solution

a.​ From the given, 𝑥 = 3 is part of the the domain 𝑥 ≤ 3. Thus,

𝑓(𝑥) = 2𝑥 + 5
𝑓(3) = 2(3) + 5
𝑓(3) = 11
Hence, 𝑓(3) exists.
b.​ In evaluating lim 𝑓(𝑥), consider the two-sided limits.
𝑥→3
lim 𝑓(𝑥) = lim 2𝑥 + 5
− −
𝑥→3 𝑥→3
= 2(3) + 5
= 11
lim 𝑓(𝑥) = lim 10 − 𝑥
+ +
𝑥→3 𝑥→3
= 10 − 3
=7
Since lim 𝑓(𝑥) ≠ lim 𝑓(𝑥), lim 𝑓(𝑥) does not exist.
− +
𝑥→3 𝑥→3 𝑥→3
Therefore, f has a jump discontinuity at 𝑥 = 3.

32
Graph

{ } is continuous at 𝑥 = 2.
3
𝑥 +1, 𝑥 <2
Determine if 𝑓(𝑥) = 5𝑥−3, 𝑥 ≥2

Solution

a.​ From the given, 𝑥 = 2 is part of the the domain 𝑥 ≥ 2. Thus,

𝑓(𝑥) = 5𝑥 − 3
𝑓(2) = 5(2) − 3
𝑓(2) = 7
Hence, 𝑓(3) exists.
b.​ In evaluating lim 𝑓(𝑥), consider the two-sided limits.
𝑥→2
3
lim 𝑓(𝑥) = lim 𝑥 + 1
− −
𝑥→2 𝑥→2
3
= (2) + 1
=9
lim 𝑓(𝑥) = lim 5𝑥 − 3
+ +
𝑥→2 𝑥→2
= 5(2) − 3
=7
Since lim 𝑓(𝑥) ≠ lim 𝑓(𝑥), lim 𝑓(𝑥) does not exist.
− +
𝑥→2 𝑥→2 𝑥→2
Therefore, f has a jump discontinuity at 𝑥 = 2.

33
 Graph

SAMPLE PROBLEMS | Infinite Discontinuity

1
Determine if 𝑓(𝑥) = 𝑥−4
is continuous at 𝑥 = 4.
Solution

a.​ Evaluate 𝑓 at 𝑥 = 4.
1
𝑓(𝑥) = 𝑥−4
1
𝑓(4) = 4−4
1
𝑓(4) = 0
, which is undefined
Since 𝑓(4) is undefined, the first continuity requirement fails and there is
no need to check the second and third conditions. Furthermore, as
𝑥 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ𝑒𝑠 4, 𝑓(𝑥) tends to ± ∞, confirming an infinite discontinuity a 𝑥 = 4.
Graph

34
 1
Determine if 𝑓(𝑥) = 2 is continuous at 𝑥 = 3.
(𝑥−3)
Solution

a.​ Evaluate 𝑓 at 𝑥 = 3.
1
𝑓(𝑥) = 2
(𝑥−3)
1
𝑓(3) = 2
(3−3)
1
𝑓(3) = 0
, which is undefined
Since 𝑓(3) is undefined, the first continuity requirement fails and there is
no need to check the second and third conditions. Furthermore, as
𝑥 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ𝑒𝑠 3, 𝑓(𝑥) tends to ± ∞, confirming an infinite discontinuity a 𝑥 = 3.
Graph

1
Determine if 𝑓(𝑥) = 𝑥+2
is continuous at 𝑥 =− 2.
Solution

a.​ Evaluate 𝑓 at 𝑥 =− 2.
1
𝑓(𝑥) = 𝑥+2
1
𝑓(− 2) = −2+2
1
𝑓(− 2) = 0
, which is undefined
Since 𝑓(− 2) is undefined, the first continuity requirement fails and there
is no need to check the second and third conditions. Furthermore, as
𝑥 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ𝑒𝑠 − 2, 𝑓(𝑥) tends to ± ∞, confirming an infinite discontinuity a
𝑥 =− 2.

35
 Graph

1
Determine if 𝑓(𝑥) = 2 is continuous at 𝑥 = 2.
(𝑥−2)
Solution

a.​ Evaluate 𝑓 at 𝑥 = 2.
1
𝑓(𝑥) = 2
(𝑥−2)
1
𝑓(2) = 2
(2−2)
1
𝑓(2) = 0
, which is undefined
Since 𝑓(2) is undefined, the first continuity requirement fails and there is
no need to check the second and third conditions. Furthermore, as
𝑥 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ𝑒𝑠 2, 𝑓(𝑥) tends to ± ∞, confirming an infinite discontinuity a 𝑥 = 2.

Graph

36
 1
Determine if 𝑓(𝑥) = 2 is continuous at 𝑥 = 4.
(𝑥−4)
Solution

a.​ Evaluate 𝑓 at 𝑥 = 4.
1
𝑓(𝑥) = 2
(𝑥−4)
1
𝑓(2) = 2
(2−2)
1
𝑓(2) = 0
, which is undefined
Since 𝑓(4) is undefined, the first continuity requirement fails and there is
no need to check the second and third conditions. Furthermore, as
𝑥 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ𝑒𝑠 4, 𝑓(𝑥) tends to ± ∞, confirming an infinite discontinuity a 𝑥 = 4.

Graph

37
Properties of Continuity One-sided Continuity

Continuity is a foundational concept A function may exhibit continuity

from one side at a point even if it is not fully
in calculus that describes functions without
continuous there. This concept is critical for
abrupt breaks, jumps, or holes. Below are analyzing piecewise functions or endpoints.
key properties governing continuity:
A function 𝑓 is said to be continuous
A polynomial function is continuous at from the left at 𝑥 = 𝑐 if
every number. 𝑓(𝑐) = lim 𝑓(𝑥)
−
𝑥→𝑐

A rational function is continuous at

every number in its domain. A function 𝑓 is said to be continuous
from the right at 𝑥 = c if
If is a positive number and 𝑓(𝑥) = 𝑛𝑥 𝑓(𝑐) = lim 𝑓(𝑥)
+
then, 𝑥→𝑐

a.​ if 𝑛 is odd, 𝑓 is continuous at

every number.
b.​ if 𝑛 is even, 𝑓 is continuous at
every nonnegative number

SAMPLE PROBLEMS | Properties of Continuity & One-sided Continuity

Determine if 𝑓(𝑥) = { 2𝑥+1,

𝑥−1 ,
𝑥 >3
𝑥 ≤1 } is continuous from the left at 𝑥 = 1.
Solution

a.​ From the given, 𝑥 = 1 is part of the the domain 𝑥 ≤ 1. Thus,

𝑓(𝑥) = 𝑥 − 1
𝑓(1) = 1 − 1
𝑓(1) = 0
Hence, 𝑓(1) exists.
b.​ In evaluating the limit from the left at 𝑥 = 1, consider lim 𝑓(𝑥).
−
𝑥→1
lim 𝑓(𝑥) = lim 𝑥 − 1
− −
𝑥→3 𝑥→3
=1−1
=0
Since 𝑓(1) = lim 𝑓(𝑥), the function f is continuous from the left at 𝑥 = 1.
−
𝑥→1

⎰ 𝑥, 𝑥 ≥4⎱
Determine if 𝑓(𝑥) = 𝑥 is continuous from the right at 𝑥 = 4.
⎱ 2
, 𝑥 <4⎰

Solution

a.​ From the given, 𝑥 = 4 is part of the the domain 𝑥 ≥ 4. Thus,

𝑓(𝑥) = 𝑥
𝑓(4) = 4
𝑓(4) = 2
Hence, 𝑓(4) exists.
b.​ In evaluating the limit from the right at 𝑥 = 4, consider lim 𝑓(𝑥).
+
𝑥→4

38
 lim 𝑓(𝑥) = lim 𝑥
+ +
𝑥→4 𝑥→4

= 4
=2
Since 𝑓(4) = lim 𝑓(𝑥), the function f is continuous from the left at 𝑥 = 4.
+
𝑥→4

Determine if 𝑓(𝑥) = { 5−𝑥,

2𝑥 ,
𝑥 >3
𝑥 ≤3 } is continuous from the left at 𝑥 = 3.
Solution

a.​ From the given, 𝑥 = 3 is part of the the domain 𝑥 ≤ 3. Thus,

𝑓(𝑥) = 2𝑥
𝑓(3) = 2(3)
𝑓(3) = 6
Hence, 𝑓(3) exists.
b.​ In evaluating the limit from the left at 𝑥 = 3, consider lim 𝑓(𝑥).
−
𝑥→3
lim 𝑓(𝑥) = lim 2𝑥
− −
𝑥→3 𝑥→3
= 2(3)
=6
Since 𝑓(3) = lim 𝑓(𝑥), the function f is continuous from the left at 𝑥 = 3.
−
𝑥→3

{ } is continuous from the right at 𝑥 = 2.

3
𝑥, 𝑥 <2
Determine if 𝑓(𝑥) = 4𝑥, 𝑥 ≥2

Solution

a.​ From the given, 𝑥 = 2 is part of the the domain 𝑥 ≥ 2. Thus,

𝑓(𝑥) = 4𝑥
𝑓(2) = 4(2)
𝑓(2) = 8
Hence, 𝑓(2) exists.
b.​ In evaluating the limit from the right at 𝑥 = 2, consider lim 𝑓(𝑥).
+
𝑥→2
lim 𝑓(𝑥) = lim 4𝑥
+ +
𝑥→2 𝑥→2
= 4(2)
=8
Since 𝑓(2) = lim 𝑓(𝑥), the function f is continuous from the left at 𝑥 = 2.
+
𝑥→2

1
⎰ , 𝑥 >5⎱
Determine if 𝑓(𝑥) = 𝑥−5
is continuous from the left at 𝑥 = 5.
⎱ 10 , 𝑥 ≤5⎰

Solution

a.​ From the given, 𝑥 = 5 is part of the the domain 𝑥 ≤ 5. Thus,

𝑓(𝑥) = 10
𝑓(3) = 10
Hence, 𝑓(3) exists.
b.​ In evaluating the limit from the left at 𝑥 = 5, consider lim 𝑓(𝑥).
−
𝑥→5

39
 lim 𝑓(𝑥) = lim 10
− −
𝑥→5 𝑥→5
= 10
Since 𝑓(5) = lim 𝑓(𝑥), the function f is continuous from the left at 𝑥 = 5.
−
𝑥→5

Continuity on an Interval

Continuity on an interval extends the

concept of continuity at a single point to
ensure a function behaves predictably
across an entire range of values. A function
𝑓(𝑥) is continuous on an interval if it meets
the criteria for continuity at every point within
that interval. This property is critical for
applications requiring seamless behavior
over domains, such as modeling physical
systems, solving equations, or applying
foundational theorems in calculus.

Closed Interval [a,b] Half-Closed Interval [a,b)

A function 𝑓 is said to be continuous on a

A function 𝑓 is said to be continuous on a
closed interval [𝒂, 𝒃] if and only if the half-closed interval [𝒂, 𝒃) if and only if the
following conditions are satisfied: following conditions are satisfied:
a.​ 𝑓 is continuous on the interval (𝑎, a.​ 𝑓 is continuous on the interval
𝑏);
(𝑎, 𝑏); and
b.​ 𝑓 is continuous from the right at
𝑥 = 𝑎; and b.​ 𝑓 is continuous from the right at
c.​ 𝑓 is continuous from the left at 𝑥 = 𝑎.
𝑥 = 𝑏.

Half-Closed Interval (a,b]

A function 𝑓 is said to be continuous on a

half-closed interval (𝒂, 𝒃] if and only if the
following conditions are satisfied:

a.​ 𝑓 is continuous on the interval

(𝑎, 𝑏); and

b.​ 𝑓 is continuous from the left at

𝑥 = 𝑏.

40
SAMPLE PROBLEMS | Closed Interval [a,b]

2
Determine if 𝑓(𝑥) = 𝑥 is continuous on [− 2, 2].
Solution

The domain of the function is D: {𝑥 ϵ ℝ} then,

a.​ 𝑓(𝑥) is continuous on the open interval (− 2, 2).
b.​ Determine if 𝑓 is continuous from the right at 𝑥 = − 2.
2
𝑓(− 2) = (− 2) = 4
2

lim 𝑥 =
𝑥 → −2
+
2
( lim 𝑥
𝑥 → −2
+ ) 2
= (− 2) = 4

Since 𝑓(− 2) = lim 𝑓(𝑥), 𝑓 is continuous from the right at 𝑥 = − 2.

+
𝑥 → −2

c.​ Determine if 𝑓 is continuous from the left at 𝑥 = − 2.

2
𝑓(2) = (− 2) = 4
2 2
lim 𝑥 = (2) = 4
−
𝑥→2

Since 𝑓(2) = lim 𝑓(𝑥), 𝑓 is continuous from the left at 𝑥 = 2.

−
𝑥→2

2
∴ Since the three conditions are satisfied, 𝑓(𝑥) = 𝑥 is continuous on [− 2, 2].

1
Determine if 𝑓(𝑥) = 2 is continuous on [1, 5].
(𝑥−3) +1
Solution

The domain of the function is D: {𝑥 ϵ ℝ} then,

a.​ 𝑓(𝑥) is continuous on the open interval (1, 5).
b.​ Determine if 𝑓 is continuous from the right at 𝑥 = 1.
1 1
𝑓(1) = 2 = 5
(1−3) +1

1 2
lim 2 = (− 2) = 4
𝑥 → −2
+ (𝑥−3) +1

Since 𝑓(− 2) = lim 𝑓(𝑥), 𝑓 is continuous from the right at 𝑥 = 1.

+
𝑥 → −2

c.​ Determine if 𝑓 is continuous from the left at 𝑥 = − 2.

2
𝑓(2) = (− 2) = 4
2 2
lim 𝑥 = (2) = 4
−
𝑥→2

Since 𝑓(2) = lim 𝑓(𝑥), 𝑓 is continuous from the left at 𝑥 = 2.

−
𝑥→2

2
∴ Since the three conditions are satisfied, 𝑓(𝑥) = 𝑥 is continuous on [− 2, 2].

41
 2
Determine if 𝑓(𝑥) = 16 − 𝑥 is continuous on [− 4, 4].
Solution

The domain of the function is D: {𝑥 ϵ ℝ | − 4 ≤ 𝑥 ≤ 4} then,

a.​ 𝑓(𝑥) is continuous on the open interval (− 4, 4).
b.​ Determine if 𝑓 is continuous from the right at 𝑥 =− 4.
2 2
𝑓(− 4) = 16 − 𝑥 = 16 − (− 4) = 0
2 2
lim 16 − 𝑥 = 16 − (− 4) = 0
𝑥 → −4

Since 𝑓(− 4) = lim 𝑓(𝑥), 𝑓 is continuous from the right at 𝑥 = − 4.

𝑥 → −4

c.​ Determine if 𝑓 is continuous from the left at 𝑥 = 4.

2 2
𝑓(4) = 16 − 𝑥 = 16 − (4) = 0
2 2
lim 16 − 𝑥 = 16 − (4) = 0
−
𝑥→4

Since 𝑓(4) = lim 𝑓(𝑥), 𝑓 is continuous from the left at 𝑥 = 4.

−
𝑥→4

2
∴ Since the three conditions are satisfied, 𝑓(𝑥) = 16 − 𝑥 is continuous on [− 4, 4].

Determine if 𝑓(𝑥) = 3𝑥 + 2 is continuous on [1, 4].

Solution

The domain of the function is D: {𝑥 ϵ ℝ} then,

a.​ 𝑓(𝑥) is continuous on the open interval (1, 4).
b.​ Determine if 𝑓 is continuous from the right at 𝑥 = 1.
𝑓(1) = 3𝑥 + 2 = 3(1) + 2 = 5
lim 3𝑥 + 2 = 3(1) + 2 = 5
+
𝑥→1

Since 𝑓(1) = lim 𝑓(𝑥), 𝑓 is continuous from the right at 𝑥 = 1.

+
𝑥→1

c.​ Determine if 𝑓 is continuous from the left at 𝑥 = 4.

𝑓(4) = 3𝑥 + 2 = 3(4) + 2 = 14
lim 3𝑥 + 2 = 3(4) + 2 = 14
−
𝑥→4

Since 𝑓(4) = lim 𝑓(𝑥), 𝑓 is continuous from the left at 𝑥 = 4.

−
𝑥→4

∴ Since the three conditions are satisfied, 𝑓(𝑥) = 3𝑥 + 2 is continuous on [1, 4].

42
 2
Determine if 𝑓(𝑥) = 𝑥 − 5𝑥 + 6 is continuous on [0, 5].
Solution

The domain of the function is D: {𝑥 ϵ ℝ} then,

a.​ 𝑓(𝑥) is continuous on the open interval (0, 5). .
b.​ Determine if 𝑓 is continuous from the right at 𝑥 = 0.
2 2
𝑓(0) = 𝑥 − 5𝑥 + 6 = (0) − 5(0) + 6 = 6
2 2
lim 𝑥 − 5𝑥 + 6 = (0) − 5(0) + 6 = 6
+
𝑥→0

Since 𝑓(0) = lim 𝑓(𝑥), 𝑓 is continuous from the right at 𝑥 = 0.

+
𝑥→0

c.​ Determine if 𝑓 is continuous from the left at 𝑥 = 5.

2 2
𝑓(5) = 𝑥 − 5𝑥 + 6 = (5) − 5(5) + 6 = 6
2 2
lim 𝑥 − 5𝑥 + 6 = (5) − 5(5) + 6 = 6
−
𝑥→5

Since 𝑓(5) = lim 𝑓(𝑥), 𝑓 is continuous from the left at 𝑥 = 5.

−
𝑥→5

2
∴ Since the three conditions are satisfied, 𝑓(𝑥) = 𝑥 − 5𝑥 + 6 is continuous on
[0, 5].

SAMPLE PROBLEMS | Half-Closed Interval [a,b)

Determine if 𝑓(𝑥) = 2𝑥 + 3 is continuous on [1, 4).

Solution

The domain of the function is D: {𝑥 ϵ ℝ} then,

a.​ 𝑓(𝑥) is continuous on the open interval (1, 4).
b.​ Determine if 𝑓 is continuous from the right at 𝑥 = 1.
𝑓(1) = 2𝑥 + 3 = 2(1) + 3 = 5
lim 2𝑥 + 3 = 2(1) + 3 = 5
+
𝑥→1

Since 𝑓(1) = lim 𝑓(𝑥), 𝑓 is continuous from the right at 𝑥 = 1.

+
𝑥→1

∴ 𝑓(𝑥) = 2𝑥 + 3 is continuous on [1, 4).

2
Determine if 𝑓(𝑥) = 𝑥 − 5𝑥 + 6 is continuous on [0, 3).
Solution

The domain of the function is D: {𝑥 ϵ ℝ} then,

a.​ 𝑓(𝑥) is continuous on the open interval (0, 3).
b.​ Determine if 𝑓 is continuous from the right at 𝑥 = 0.
2 2
𝑓(0) = 𝑥 − 5𝑥 + 6 = (0) − 5(0) + 6 = 6
43
 2 2
lim 𝑥 − 5𝑥 + 6 = (0) − 5(0) + 6 = 6
+
𝑥→0

Since 𝑓(0) = lim 𝑓(𝑥), 𝑓 is continuous from the right at 𝑥 = 0.

+
𝑥→0

2
∴ 𝑓(𝑥) = 𝑥 − 5𝑥 + 6 is continuous on [0, 3).

Determine if 𝑓(𝑥) = |𝑥 − 2| is continuous on [0, 5).

Solution

The domain of the function is D: {𝑥 ϵ ℝ} then,

a.​ 𝑓(𝑥) is continuous on the open interval (0, 5).
b.​ Determine if 𝑓 is continuous from the right at 𝑥 = 0.
𝑓(0) = |𝑥 − 2| = |0 − 2| = 2
lim |𝑥 − 2| = |0 − 2| = 2
+
𝑥→0

Since 𝑓(0) = lim 𝑓(𝑥), 𝑓 is continuous from the right at 𝑥 = 0.

+
𝑥→0

∴ 𝑓(𝑥) = |𝑥 − 2| is continuous on [0, 5).

Determine if 𝑓(𝑥) = 𝑥 − 1 is continuous on [1, 5).

Solution

The domain of the function is D: {𝑥 ϵ ℝ | 𝑥 ≥ 1} then,

a.​ 𝑓(𝑥) is continuous on the open interval (1, 5).
b.​ Determine if 𝑓 is continuous from the right at 𝑥 = 1.

𝑓(1) = 𝑥 − 1 = 1 − 1 = 0

lim 𝑥 − 1= 1 − 1= 0
+
𝑥→1

Since 𝑓(1) = lim 𝑓(𝑥), 𝑓 is continuous from the right at 𝑥 = 1.

+
𝑥→1

∴ 𝑓(𝑥) = 𝑥 − 1 is continuous on [1, 5).

1
Determine if 𝑓(𝑥) = 𝑥−3
is continuous on [0, 3).
Solution

The domain of the function is D: {𝑥 ϵ ℝ | 𝑥 ≠ 3} then,

a.​ 𝑓(𝑥) is continuous on the open interval (0, 3).
b.​ Determine if 𝑓 is continuous from the right at 𝑥 = 0.
1 1 1
𝑓(0) = 𝑥−3
= 0−3
=− 3

1 1 1
lim 𝑥−3
= 0−3
=− 3
+
𝑥→0

Since 𝑓(0) = lim 𝑓(𝑥), 𝑓 is continuous from the right at 𝑥 = 0.

+
𝑥→0

44
 1
∴ 𝑓(𝑥) = 𝑥−3
is continuous on [0, 3).

SAMPLE PROBLEMS | Half-Closed Interval (a,b]

Determine if 𝑓(𝑥) = 4𝑥 − 7 is continuous on (2, 5].

Solution

The domain of the function is D: {𝑥 ϵ ℝ } then,

a.​ 𝑓(𝑥) is continuous on the open interval (2, 5).
b.​ Determine if 𝑓 is continuous from the left at 𝑥 = 5.
𝑓(5) = 4𝑥 − 7 = 4(5) − 7 = 13
lim 4𝑥 − 7 = 4(5) − 7 = 13
−
𝑥→5

Since 𝑓(5) = lim 𝑓(𝑥), 𝑓 is continuous from the left at 𝑥 = 5.

−
𝑥→5

∴ 𝑓(𝑥) = 4𝑥 − 7 is continuous on (2, 5].

3
Determine if 𝑓(𝑥) = 𝑥 − 2𝑥 + 1 is continuous on (− 1, 3].
Solution

The domain of the function is D: {𝑥 ϵ ℝ} then,

a.​ 𝑓(𝑥) is continuous on the open interval (− 1, 3).
b.​ Determine if 𝑓 is continuous from the left at 𝑥 = 3.
3 3
𝑓(3) = 𝑥 − 2𝑥 + 1 = (3) − 2(3) + 1 = 22
3 3
lim (𝑥 − 2𝑥 + 1) = (3) − 2(3) + 1 = 22
−
𝑥→3

Since 𝑓(3) = lim 𝑓(𝑥), 𝑓 is continuous from the left at 𝑥 = 3.

−
𝑥→3

3
∴ 𝑓(𝑥) = 𝑥 − 2𝑥 + 1 is continuous on (− 1, 3].

Determine if 𝑓(𝑥) = |𝑥 + 1| is continuous on (− 3, 2].

Solution

The domain of the function is D: {𝑥 ϵ ℝ} then,

a.​ 𝑓(𝑥) is continuous on the open interval (− 3, 2).
b.​ Determine if 𝑓 is continuous from the left at 𝑥 = 2.
𝑓(2) = |𝑥 + 1| = |2 + 1| = 3
lim |𝑥 + 1| = |2 + 1| = 3
−
𝑥→2

Since 𝑓(2) = lim 𝑓(𝑥), 𝑓 is continuous from the left at 𝑥 = 2.

−
𝑥→2

∴ 𝑓(𝑥) = |𝑥 + 1| is continuous on (− 3, 2].

Determine if 𝑓(𝑥) = 𝑥 − 1 is continuous on (1, 4].

45
 Solution

The domain of the function is D: {𝑥 ϵ ℝ | 𝑥 ≥ 1} then,

a.​ 𝑓(𝑥) is continuous on the open interval (1, 4).
b.​ Determine if 𝑓 is continuous from the right at 𝑥 = 4.

𝑓(4) = 𝑥 − 1 = 4 − 1 = 3

lim 𝑥 − 1= 4 − 1= 3
−
𝑥→4

Since 𝑓(4) = lim 𝑓(𝑥), 𝑓 is continuous from the right at 𝑥 = 4.

−
𝑥→4

∴ 𝑓(𝑥) = 𝑥 − 1 is continuous on (1, 4].

1
Determine if 𝑓(𝑥) = 𝑥−5
is continuous on (0, 4].
Solution

The domain of the function is D: {𝑥 ϵ ℝ | 𝑥 ≠ 5} then,

a.​ 𝑓(𝑥) is continuous on the open interval (0, 4).
b.​ Determine if 𝑓 is continuous from the right at 𝑥 = 4.
1 1
𝑓(4) = 𝑥−5
= 4−5
=− 1
1 1
lim 𝑥−5
= 4−5
=− 1
−
𝑥→4

Since 𝑓(4) = lim 𝑓(𝑥), 𝑓 is continuous from the right at 𝑥 = 4.

−
𝑥→4

1
∴ 𝑓(𝑥) = 𝑥−5
is continuous on (0, 4].

46
 DERIVATIVES

In this chapter, you should learn and meet the

following objectives:

Illustrate the tangent lines to a

function’s graph.

Apply the derivative definition to find a In industrial piping systems, PID (Proportional-Integral-Derivative)
function’s derivative at a point. controllers are responsible for regulating parameters such as flow
rate, pressure, and temperature. The derivative (𝐷) term in a PID
controller helps improve system stability by predicting future errors
Relate derivatives to slopes of tangent and applying corrective action before the error becomes too large.
It does this by calculating the rate of change of the error over time,
lines. allowing the system to respond more proactively to disturbances.

Use differentiation rules to compute

Tangent Line
derivatives of algebraic, exponential,
logarithmic, trigonometric, and inverse A tangent line to a curve at a specific
trigonometric functions. point is a straight line that touches the curve at
that point without intersecting it. It aligns
Introduction precisely with the slope of the curve at that
Calculus is the branch of mathematics location. The slope of this tangent line
that studies how things change. At its heart is corresponds to the derivative of the function at
the concept of the derivative, which quantifies that point, signifying the instantaneous rate of
the instantaneous rate of change of a function change. This geometric concept is essential in
with respect to its independent variable. In calculus, bridging visual intuition with analytical
practical terms, if you have a function that precision.
describes a physical quantity (like position,
temperature, or cost) as it varies with time or
another variable, the derivative tells you how
quickly that quantity is changing at any given
moment.

47
Secant Line
A secant line is a straight line that
intersects a curve at two distinct points,
providing a measure of the average rate of
change between them. In calculus, the slope of
a secant line, calculated as
𝑓(𝑥)−𝑓(𝑎)
𝑥−𝑎
, approximates the behavior of the
curve over an interval. As the two points
converge, the secant line transitions into the
tangent line at 𝑥 = 𝑎, with its slope
approaching the derivative 𝑓'(𝑎). This process
underpins the foundational principle of
derivatives, linking discrete average rates to The image shows a curve with two points: (𝑎, 𝑓(𝑎)) and
(𝑥, 𝑓(𝑥)). Straight orange line passes through these
instantaneous rates of change and enabling points–this is the secant line, representing the average rate
precise analysis of dynamic systems in of change between the two points. The slope of this line is
𝑓(𝑥)−𝑓(𝑎)
mathematics. calculated using the formula 𝑥−𝑎
.

Limit Definition of the Slope of a

Tangent Line to a Given Curve
The slope of a tangent line at a point
on a curve is the limit of the slope of a secant
line as the second point moves closer to the
first. It is given by:
𝑓(𝑥)−𝑓(𝑥0)
𝑀𝑡 = lim 𝑥−𝑥0
𝑥 → 𝑥0
This shows how the function changes
at exactly 𝑥0

SAMPLE PROBLEMS | Limit Definition of the Slope of a Tangent Line to a Given Curve

2
Find the general form of the equation of the tangent line to 𝑓(𝑥) = 𝑥 + 1 at the point
(1, 2).
Solution

Given the following,

(𝑥0, 𝑦0) = (1, 2)
2
𝑓(𝑥) = 𝑥 + 1
Then,
𝑓(𝑥0) = 𝑓(1)
2
= 1 +1
𝑓(𝑥0) = 2

Substitute the values in the equation,

48
 𝑓(𝑥)−𝑓(𝑥0)
𝑀𝑡 = lim 𝑥−𝑥0
𝑥 → 𝑥0
2
(𝑥 +1)−(2)
= lim 𝑥−1
𝑥→1
2
𝑥 −1
= lim 𝑥−1
𝑥→1
(𝑥−1)(𝑥+1)
= lim 𝑥−1
𝑥→1
= lim (𝑥 + 1) = lim 𝑥 + lim 1
𝑥→1 𝑥→1 𝑥→1
𝑀𝑡 = 1 + 1 = 2

2
Hence, the slope of the line tangent to 𝑓(𝑥) = 𝑥 + 1 at the point (1, 2) is 2.

To determine the equation of the tangent line to function 𝑓 , we use the

point-slope form of an equation which is 𝑦 − 𝑦0 = 𝑀𝑡(𝑥 − 𝑥0).
Substitute the values,
𝑦 − (2) = 2(𝑥 − 1)
𝑦 − 2 = 2𝑥 − 2
− 2𝑥 + 𝑦 = 0
2𝑥 − 𝑦 = 0

​ ​ Therefore, the general form of the equation of the tangent line to

2
𝑓(𝑥) = 𝑥 + 1 at the point (1, 2) is 2𝑥 − 𝑦 = 0.

2
Find the general form of the equation of the tangent line to 𝑓(𝑥) = 𝑥 − 4 at the point
(3, 5).
Solution

Given the following,

(𝑥0, 𝑦0) = (3, 5)
2
𝑓(𝑥) = 𝑥 − 4
Then,
𝑓(𝑥0) = 𝑓(3)
2
= (3) − 4
𝑓(𝑥0) = 5

Substitute the values in the equation,

𝑓(𝑥)−𝑓(𝑥0)
𝑀𝑡 = lim 𝑥−𝑥0
𝑥 → 𝑥0
2
(𝑥 −4)−(5)
= lim 𝑥−3
𝑥→3
2
𝑥 −9
= lim 𝑥−3
𝑥→3
(𝑥−3)(𝑥+3)
= lim 𝑥−3
𝑥→3
= lim (𝑥 + 3) = lim 𝑥 + lim 3
𝑥→3 𝑥→3 𝑥→3

49
 𝑀𝑡 = 3 + 3 = 6

2
Hence, the slope of the line tangent to 𝑓(𝑥) = 𝑥 − 4 at the point (3, 5) is 6.

To determine the equation of the tangent line to function 𝑓 , we use the

point-slope form of an equation which is 𝑦 − 𝑦0 = 𝑀𝑡(𝑥 − 𝑥0).
Substitute the values,
𝑦 − (5) = 6(𝑥 − 3)
𝑦 − 5 = 6𝑥 − 18
− 6𝑥 + 𝑦 + 13 = 0
6𝑥 − 𝑦 − 13 = 0

​ ​ Therefore, the general form of the equation of the tangent line to

2
𝑓(𝑥) = 𝑥 − 4 at the point (3, 5) is 6𝑥 − 𝑦 − 13 = 0.

2
Find the general form of the equation of the tangent line to 𝑓(𝑥) = 2𝑥 at the point (2, 8).
Solution

Given the following,

(𝑥0, 𝑦0) = (2, 8)
2
𝑓(𝑥) = 2𝑥
Then,
𝑓(𝑥0) = 𝑓(2)
2
= 2(2)
𝑓(𝑥0) = 8

Substitute the values in the equation,

𝑓(𝑥)−𝑓(𝑥0)
𝑀𝑡 = lim 𝑥−𝑥0
𝑥 → 𝑥0
2
(2𝑥 )−(8)
= lim 𝑥−2
𝑥→2
2
2(𝑥 −4)
= lim 𝑥−2
𝑥→2
2(𝑥−2)(𝑥+2)
= lim 𝑥−2
𝑥→2
= lim 2𝑥 + 4 = 2 lim 𝑥 + lim 4
𝑥→2 𝑥→2 𝑥→2
𝑀𝑡 = 2(2) + 4 = 8

2
Hence, the slope of the line tangent to 𝑓(𝑥) = 2𝑥 at the point (2, 8) is 8.

To determine the equation of the tangent line to function 𝑓 , we use the

point-slope form of an equation which is 𝑦 − 𝑦0 = 𝑀𝑡(𝑥 − 𝑥0).
Substitute the values,
𝑦 − (8) = 8(𝑥 − 2)
𝑦 − 8 = 8𝑥 − 16
− 8𝑥 + 𝑦 + 8 = 0

50
 8𝑥 − 𝑦 − 8 = 0

​ ​ Therefore, the general form of the equation of the tangent line to

2
𝑓(𝑥) = 2𝑥 at the point (2, 8) is 8𝑥 − 𝑦 − 8 = 0.

2
Find the general form of the equation of the tangent line to 𝑓(𝑥) = 𝑥 + 2𝑥 at the point
(1, 3).
Solution

Given the following,

(𝑥0, 𝑦0) = (1, 3)
2
𝑓(𝑥) = 𝑥 + 2𝑥
Then,
𝑓(𝑥0) = 𝑓(1)
2
= (1) + 2(1)
𝑓(𝑥0) = 3

Substitute the values in the equation,

𝑓(𝑥)−𝑓(𝑥0)
𝑀𝑡 = lim 𝑥−𝑥0
𝑥 → 𝑥0
2
(𝑥 +2𝑥)−(3)
= lim 𝑥−1
𝑥→1
2
𝑥 +2𝑥−3
= lim 𝑥−1
𝑥→1
(𝑥−1)(𝑥+3)
= lim 𝑥−1
𝑥→1
= lim (𝑥 + 3) = lim 𝑥 + lim 3
𝑥→1 𝑥→1 𝑥→1
𝑀𝑡 = 1 + 3 = 4

2
Hence, the slope of the line tangent to 𝑓(𝑥) = 𝑥 + 2𝑥 at the point (1, 3) is 4.

To determine the equation of the tangent line to function 𝑓 , we use the

point-slope form of an equation which is 𝑦 − 𝑦0 = 𝑀𝑡(𝑥 − 𝑥0).
Substitute the values,
𝑦 − (3) = 4(𝑥 − 1)
𝑦 − 3 = 4𝑥 − 4
− 4𝑥 + 𝑦 + 1 = 0
4𝑥 − 𝑦 − 1 = 0

​ ​ Therefore, the general form of the equation of the tangent line to

2
𝑓(𝑥) = 𝑥 + 2𝑥 at the point (1, 3) is 4𝑥 − 𝑦 − 1 = 0.

51
 2
Find the general form of the equation of the tangent line to 𝑓(𝑥) = 𝑥 − 𝑥 at the point
(2, 2).
Solution

Given the following,

(𝑥0, 𝑦0) = (2, 2)
2
𝑓(𝑥) = 𝑥 − 𝑥
Then,
𝑓(𝑥0) = 𝑓(2)
2
= 2 −2
𝑓(𝑥0) = 2

Substitute the values in the equation,

𝑓(𝑥)−𝑓(𝑥0)
𝑀𝑡 = lim 𝑥−𝑥0
𝑥 → 𝑥0
2
(𝑥 −𝑥)−(2)
= lim 𝑥−2
𝑥→2
2
𝑥 −𝑥−2
= lim 𝑥−2
𝑥→2
(𝑥−2)(𝑥+1)
= lim 𝑥−2
𝑥→2
= lim (𝑥 + 1) = lim 𝑥 + lim 1
𝑥→2 𝑥→2 𝑥→2
𝑀𝑡 = 2 + 1 = 3

2
Hence, the slope of the line tangent to 𝑓(𝑥) = 𝑥 − 𝑥 at the point (2, 2) is 3.

To determine the equation of the tangent line to function 𝑓 , we use the

point-slope form of an equation which is 𝑦 − 𝑦0 = 𝑀𝑡(𝑥 − 𝑥0).
Substitute the values,
𝑦 − (2) = 3(𝑥 − 2)
𝑦 − 2 = 3𝑥 − 6
− 3𝑥 + 𝑦 + 4 = 0
3𝑥 − 𝑦 − 4 = 0

​ ​ Therefore, the general form of the equation of the tangent line to

2
𝑓(𝑥) = 𝑥 − 𝑥 at the point (2, 2) is 3𝑥 − 𝑦 − 4 = 0.

52
The Limit Definition

As mentioned earlier, a derivative the tangent line to the graph of 𝑓 at 𝑥 = a.

quantifies the instantaneous rate of change This provides the limit definition of the
of a function with respect to its independent derivative. This definition can also be written
variable. Formally, if 𝑓(𝑥) is a function as
defined on an interval and a is a point in that
interval, the derivative of 𝑓 at a is defined as 𝑓'(𝑥) = lim
𝑓(𝑥+∆𝑥)−𝑓(𝑥)
.
∆𝑥
∆𝑥 → 0
𝑓(𝑎+ℎ)−𝑓(𝑎)
𝑓'(𝑎) = lim ℎ
,
ℎ→0 The derivative of a function at x =
x0 represents the function's instantaneous
provided this limit exists (Stewart, rate of change at that specific point. It
2002). This expression is known as the corresponds to the slope of the tangent line
difference quotient, and as ℎ approaches
to the graph of the function at x = x0.
zero, the quotient approaches the slope of

SAMPLE PROBLEMS | Differentiation using the Limit Definition

Find the derivative of 𝑓(𝑥) = 3𝑥.

​
Solution
𝑓(𝑥+∆𝑥) − 𝑓(𝑥)
𝑓'(𝑥) = lim ∆𝑥
∆𝑥 → 0
3(𝑥+∆𝑥) − 3(𝑥)
​ = lim ∆𝑥
∆𝑥 → 0
3𝑥+3∆𝑥 − 3𝑥
​ = lim ∆𝑥
∆𝑥 → 0
3∆𝑥
​ = lim ∆𝑥
∆𝑥 → 0
​ = lim 3
∆𝑥 → 0
​ =3

2
Find the derivative of 𝑓(𝑥) = 7𝑥 .
​
Solution
𝑓(𝑥+∆𝑥) − 𝑓(𝑥)
𝑓'(𝑥) = lim ∆𝑥
∆𝑥 → 0
2 2
7(𝑥+∆𝑥) − 7(𝑥)
​ = lim ∆𝑥
∆𝑥 → 0
2 2 2
7(𝑥 +2𝑥∆𝑥+∆𝑥 )− 7𝑥
​ = lim ∆𝑥
∆𝑥 → 0
2 2 2
7𝑥 +14𝑥∆𝑥+7∆𝑥 − 7𝑥
​ = lim ∆𝑥
∆𝑥 → 0

53
 2
14𝑥∆𝑥+7∆𝑥
​ = lim ∆𝑥
∆𝑥 → 0
∆𝑥(14𝑥+7∆𝑥)
​ = lim ∆𝑥
∆𝑥 → 0
​ = lim (14𝑥 + 7∆𝑥)
∆𝑥 → 0
​ = 14𝑥 + 7(0)
​ = 14x

Find the derivative of 𝑓(𝑥) = 𝑥.

​
Solution
𝑓(𝑥+∆𝑥) − 𝑓(𝑥)
𝑓'(𝑥) = lim ∆𝑥
∆𝑥 → 0
(𝑥+∆𝑥) − 𝑥
​ = lim ∆𝑥
∆𝑥 → 0
(𝑥+∆𝑥) − 𝑥 (𝑥+∆𝑥) + 𝑥
​ = lim ∆𝑥
·
(𝑥+∆𝑥) + 𝑥
∆𝑥 → 0
(𝑥+∆𝑥) − (𝑥)
​ = lim
∆𝑥( (𝑥+∆𝑥) + 𝑥)
∆𝑥 → 0
∆𝑥
​ = lim
∆𝑥( (𝑥+∆𝑥) + 𝑥)
∆𝑥 → 0
1
​ = lim
(𝑥+∆𝑥) + 𝑥
∆𝑥 → 0
1
​ =
(𝑥+(0)) + 𝑥
1
​ =
(𝑥) + 𝑥
1
​ =
2 𝑥
1 𝑥
​ = ·
2 𝑥 𝑥
𝑥
​ = 2𝑥

2
Find the derivative of 𝑤(𝑥) = 𝑥 + 2𝑥 + 1.
​
Solution
𝑓(𝑥+∆𝑥) − 𝑓(𝑥)
​ 𝑤'(𝑥) = lim ∆𝑥
∆𝑥 → 0
2 2
(𝑥+∆𝑥) +2(𝑥+∆𝑥)+1 − [𝑥 +2𝑥+1]
​ = lim ∆𝑥
∆𝑥 → 0
2 2 2
𝑥 +2𝑥∆𝑥+∆𝑥 +2𝑥+2∆𝑥+1 − 𝑥 −2𝑥−1
​ = lim ∆𝑥
∆𝑥 → 0
2
2𝑥∆𝑥+∆𝑥 +2∆𝑥
​ = lim ∆𝑥
∆𝑥 → 0
∆𝑥(2𝑥+∆𝑥+2)
​ = lim ∆𝑥
∆𝑥 → 0
​ = lim (2𝑥 + ∆𝑥 + 2)
∆𝑥 → 0
​ = 2𝑥 + (0) + 2
54
 ​ = 2𝑥 + 2

Find the derivative of 𝑓(𝑥) = 𝑥 3 .

​
Solution
𝑓(𝑥+∆𝑥) − 𝑓(𝑥)
𝑓'(𝑥) = lim ∆𝑥
∆𝑥 → 0
1 1

(𝑥+∆𝑥) 3 − 𝑥 3
= lim ∆𝑥
∆𝑥 → 0
1 1 2 1 1 2

(𝑥+∆𝑥) 3 − (𝑥) 3 (𝑥+∆𝑥) 3 +(𝑥+∆𝑥) 3 (𝑥) 3 +(𝑥) 3

= lim ∆𝑥
⋅ 2 1 1 2

∆𝑥 → 0 (𝑥+∆𝑥) 3 +(𝑥+∆𝑥) 3 (𝑥) 3 +(𝑥) 3

2 1 1 2 2 1 1 2

(𝑥+∆𝑥)+(𝑥+∆𝑥) 3 (𝑥) 3 +(𝑥+∆𝑥) 3 (𝑥) 3 −(𝑥+∆𝑥) 3 (𝑥) 3 −(𝑥+∆𝑥) 3 (𝑥) 3 −(𝑥)

= lim 2 1 1 2

∆𝑥 → 0
3 3 3 3
∆𝑥((𝑥+∆𝑥) +(𝑥+∆𝑥) (𝑥) +(𝑥) )
(𝑥+∆𝑥)−(𝑥)
= lim 2 1 1 2

∆𝑥 → 0 ∆𝑥((𝑥+∆𝑥) 3 +(𝑥+∆𝑥) 3 (𝑥) 3 +(𝑥) 3 )

∆𝑥
= lim 2 1 1 2

∆𝑥 → 0 ∆𝑥((𝑥+∆𝑥) 3 +(𝑥+∆𝑥) 3 (𝑥) 3 +(𝑥) 3 )

1
= lim 2 1 1 2

∆𝑥 → 0 (𝑥+∆𝑥) +(𝑥+∆𝑥) 3 (𝑥) 3 +(𝑥) 3

3

1
= 2 1 1 2

(𝑥) 3 +(𝑥) 3 (𝑥) 3 +(𝑥) 3

1
= 2 2 2

(𝑥) 3 +(𝑥) 3 +(𝑥) 3

1
= 2

3(𝑥) 3
1

1 (𝑥) 3
= 2 ⋅ 1

3(𝑥) 3 (𝑥) 3
1
3
𝑥
= 3𝑥

Differentiation Rules
derivative of the entire expression can be
Differentiation rules in calculus are a found by separately differentiating each
set of systematic guidelines that allow you to component and then combining those results
determine the derivative, or the rate at which a accordingly. Similarly, if a function is multiplied
function changes, without having to revert to by a constant, that constant can be factored
the limit definition every time. These rules out of the differentiation process. For products
enable us to break down complex functions or quotients of functions, there are dedicated
into simpler parts whose derivatives are easier rules that offer a step-by-step method to
to compute. For instance, there is a rule that handle the interaction between the functions
tells us the derivative of a constant value is involved. Finally, when one function is nested
zero, which makes sense because a constant within another, a special rule (often called the
function does not change. Another chain rule) allows us to differentiate the
fundamental rule applies to functions where composite function by relating the rate of
the variable is raised to a power; this rule change of the outer function to that of the inner
provides an efficient way to differentiate such function. However, the chain rule will not be
expressions by adjusting the exponent in a discussed since it will be beyond the scope of
predictable manner. Additionally, when this performance task. Together, these rules
functions are combined through addition or form the foundation for efficiently computing
subtraction, specific rules state that the derivatives in a wide variety of contexts.
55
 .

Constant Rule

The constant rule of differentiation

states that the derivative of a constant function
is zero. In other words, if a function is defined
by

𝑓(𝑥) = 𝑐,

where c is a constant (a fixed real

number independent of x), then its derivative
with respect to x is In this given graph, we can see that the derivative or the slope is
equal to 0 when f(x) is just a constant (In this case, f(x)=3) since it
is merely a horizontal line.
𝑓'(𝑥) = 0,

This makes intuitive sense because a

constant function has a horizontal graph (a flat
line), which has a slope of zero.

If you want to see if this is true, you

must use the limit definition of the derivative in
order to see that constants do in fact have a
slope or derivative equal to zero (0).

SAMPLE PROBLEMS | Differentiation using the Constant Rule

Find the derivative of 𝑓(𝑥) = 𝑒.

​
Solution
​ 𝑓'(𝑥) = 0

Find the derivative of 𝑓(𝑥) = 𝑙𝑛(10).

​
Solution
​ 𝑓'(𝑥) = 0

Find the derivative of 𝑓(𝑥) = 10302193021921 + 334.

​
Solution
​ 𝑓'(𝑥) = 0
56
 3
Find the derivative of 𝑓(𝑥) = 𝑒 .
​
Solution
​ 𝑓'(𝑥) = 0

Find the derivative of 𝑓(𝑥) = π.

​
Solution
​ 𝑓'(𝑥) = 0

The Power Rule

The power rule is a fundamental tool in bypass the more laborious limit definition of the
differentiation that provides a quick and derivative. It applies not only to positive integer
straightforward way to compute the derivative exponents but can also be extended to
of functions where the variable is raised to a negative and fractional exponents (Math
power. Essentially, when you have a function LibreTexts, 2025). The power rule can be
in which x is raised to any exponent, the rule expressed as
tells you that you can differentiate it by
multiplying by the exponent and then reducing 𝑛−1
𝑓'(𝑥) = 𝑛𝑥 .
the exponent by one. This rule significantly
simplifies the process because it allows you to

SAMPLE PROBLEMS | Differentiation Using the Power Rule

2
Find the derivative of 𝑓(𝑥) = 𝑥 .
​
Solution
2−1
​ 𝑓'(𝑥) = 2 · 𝑥
​ ​ ​ ​ = 2𝑥

348593285
Find the derivative of 𝑓(𝑥) = 𝑥 .
​
Solution
348593285−1
𝑓'(𝑥) = 348593285 · 𝑥
348593284
= 348593285𝑥

1
Find the derivative of𝑓(𝑥) = 2 .
𝑥
​
Solution
−2
𝑓(𝑥) = 𝑥
−2−1
𝑓'(𝑥) =− 2 · 𝑥

57
 −3
=− 2𝑥

1
Find the derivative of 𝑓(𝑥) = 12 .
𝑥
​
Solution
−12
𝑓(𝑥) = 𝑥
−12−1
𝑓'(𝑥) =− 12 · 𝑥
−13
=− 12𝑥

Find the derivative of 𝑓(𝑥) = 𝑥.

​
Solution
1
1 2
𝑓(𝑥) = 2
·𝑥 .
1
1 2
−1
𝑓'(𝑥) = 2
·𝑥
−1
1 2
= 2
·𝑥
1
=
2 𝑥
1 𝑥
= ·
2 𝑥 𝑥
𝑥
= 2𝑥

The Constant Multiple Rule

The constant multiple rule states that if differentiate a function that has been multiplied
you multiply a differentiable function by a by a constant, you first differentiate the
constant, the derivative of that product is function itself, and then multiply the result by
simply the constant times the derivative of the that constant (Math LibreTexts, 2025; Fiveable,
function. In other words, the constant can be 2025).
“factored out” of the differentiation process ​ The constant multiple rule of
because it does not change as the variable differentiation can be expressed as
changes. This rule streamlines the
𝑑
differentiation of functions that are scaled by a 𝑑𝑥
[𝑎𝑓(𝑥)] = 𝑎𝑓'(𝑥).
constant, ensuring that the process focuses
solely on how the variable-dependent part of
the function changes. Essentially, when you

58
SAMPLE PROBLEMS | Differentiation using the Constant Rule

3 4
Find the derivative of 𝑔(𝑥) = 4
𝑥.
​
Solution
3 4−1
​ 𝑔'(𝑥) = 4 · 4
𝑥
3
​ = 3𝑥

1 6
Find the derivative of 𝑚(𝑥) = 3
𝑥.
​
Solution
1 6−1
​ 𝑚'(𝑥) = 6 · 3
𝑥
5
= 2𝑥

3
Find the derivative of ℎ(𝑥) = 2𝑥 .
​
Solution
3−1
​ ℎ'(𝑥) = 3 · 2𝑥
2
​ ​ ​ ​ = 6𝑥

45
Find the derivative of 𝑞(𝑥) = 21𝑥
​
Solution
45−1
​ 𝑞'(𝑥) = 45 · 21𝑥
44
​ ​ ​ ​ = 945𝑥

343
Find the derivative of 𝑟(𝑥) =− 11𝑥

​
Solution
343−1
​ 𝑟'(𝑥) = 343 · (− 11𝑥 )
342
=− 3773𝑥

59
The Sum and Difference Rule

In calculus, the sum and difference 𝑑

[𝑓(𝑥) ± 𝑔(𝑥)] =
𝑑
𝑓(𝑥) ±
𝑑
𝑔(𝑥).
𝑑𝑥 𝑑𝑥 𝑑𝑥
rule states that the derivative of the sum (or
The sum and difference rule simplifies
difference) of two functions is the sum (or
differentiation by allowing us to differentiate
difference) of their derivatives. Mathematically,
each term separately and then combine the
this is written as:
results. This is one of the fundamental rules
that make differentiation easier when dealing
with multiple terms

SAMPLE PROBLEMS | Differentiation using the Sum and Difference Rule

𝑑 2 2
Find the derivative of 𝑑𝑥
(3𝑥 − 2 ).
𝑥
​
Solution
𝑑 2 2 𝑑 2 −2
𝑑𝑥
(3𝑥 − 2 )= 𝑑𝑥
(3𝑥 ) − (2𝑥 )
𝑥
2−1 −3
​ = 2 · 3𝑥 − (− 2) · 2𝑥 ​ ​
−3
= 6𝑥 + 4𝑥 ​
4
= 6𝑥 + 3
𝑥

𝑑 8
Find the derivative of 𝑑𝑥
(6𝑥 + 7𝑥).
​
Solution
𝑑 8 𝑑 8 𝑑
𝑑𝑥
(6𝑥 + 7𝑥) = 𝑑𝑥
(6𝑥 ) + 𝑑𝑥
(7𝑥)
8−1 1−1
​ ​ = 8 · 6𝑥 + 7𝑥 ​ ​ ​
7
​ = 48𝑥 +7

𝑑 14 47
Find the derivative of 𝑑𝑥
(− 19𝑥 + 7𝑥 ).
​
Solution
𝑑 14 47 𝑑 14 𝑑 47
𝑑𝑥
(− 19𝑥 + 7𝑥 ) = 𝑑𝑥
(− 19𝑥 ) + 𝑑𝑥
(7𝑥 )
14−1 47−1
= 14 · (− 19)𝑥 + 47 · 7𝑥 ​
13 46
=− 266𝑥 + 329𝑥

60
 𝑑 8 13
Find the derivative of 𝑑𝑥
(𝑥 − 7𝑥 ).
​
Solution
𝑑 8 13 𝑑 8 𝑑 13
​ ​ ​ 𝑑𝑥
(𝑥 − 7𝑥 ) = 𝑑𝑥
(𝑥 ) + 𝑑𝑥
(7𝑥 )
8−1 13−1
​ = 8 ·𝑥 + 13 · 7𝑥 ​ ​ ​
7 12
​​ = 8𝑥 + 91𝑥

𝑑 1 𝑥
Find the derivative of 𝑑𝑥
( 9 − 98
).
𝑥
​
Solution
𝑑 1 𝑥 𝑑 1 𝑑 𝑥
​ ​ 𝑑𝑥
( 9 − 98
)= 𝑑𝑥
( 9 )− 𝑑𝑥
( 98 )
𝑥 𝑥
1

𝑑 −9 𝑑 𝑥2
​ ​ = 𝑑𝑥
(𝑥 ) − 𝑑𝑥
( 98 )
1
−1
−9−1 1 𝑥2
​ =− 9 · 𝑥 − 2
· 98
​ ​ ​
−1

−10 𝑥2

​ =− 9𝑥 − 196
−9 1
​ = 10 −
𝑥 196 𝑥
−9 1 𝑥
​ = 10 − ·
𝑥 196 𝑥 𝑥
−9 𝑥
​ = 10 − 196𝑥
𝑥

The Product Rule

When dealing with functions that are variable. If we have two such quantities, say
products of two differentiable functions, we 𝑓(𝑥) and 𝑔(𝑥), their product represents a new
cannot simply differentiate each function combined quantity. If both of these functions
separately and multiply the results. Instead, we change with respect to x, we need to
use the Product Rule, which provides a determine how their product changes. The
structured way to differentiate a function that is Product Rule considers two key contributions
the product of two functions. The Product Rule to this change:
states that if we have two differentiable The change in the first function while keeping
functions, 𝑓(𝑥) and 𝑔(𝑥), then the derivative of the second function fixed: this corresponds to
their product is given by: 𝑓(𝑥) remaining as it is while 𝑔(𝑥) changes at a
rate given by 𝑔'(𝑥). The change in the second
𝑑 𝑑 𝑑
𝑑𝑥
[𝑓(𝑥)𝑔(𝑥)] = 𝑓(𝑥) 𝑑𝑥
𝑔(𝑥) + 𝑔(𝑥) 𝑑𝑥
𝑓(𝑥) function while keeping the first function fixed:
here, 𝑔(𝑥) remains constant while 𝑓(𝑥) changes
To understand why this rule works, at a rate given by 𝑓'(𝑥).
let's think of a function as representing some
measurable quantity that depends on another
61
SAMPLE PROBLEMS | Differentiation using the Product Rule

Find the derivative of ℎ(𝑥) = (𝑥 − 5)(3 − 4𝑥).

​
Solution
𝑑 𝑑
ℎ'(𝑥) = [ 𝑑𝑥 (𝑥 − 5)](3 − 4𝑥) + (𝑥 − 5)[ 𝑑𝑥 (3 − 4𝑥)]
1−1 1−1
= (𝑥 − 0)(3 − 4𝑥) + (𝑥 − 5)(0 − 4𝑥 )
= (1)(3 − 4𝑥) + (𝑥 − 5)(− 4)​ ​
= 3 − 4𝑥 − 4𝑥 + 20
= − 8𝑥 + 23

2
Find the derivative of 𝑓(𝑥) = (2𝑥 − 1)(3𝑥 + 5𝑥 − 2).
​
Solution
𝑑 2 𝑑 2
𝑓'(𝑥) = [ 𝑑𝑥 (2𝑥 − 1)](3𝑥 + 5𝑥 − 2) + (2𝑥 − 1)[ 𝑑𝑥 (3𝑥
+ 5𝑥 − 2)]
1−1 2 2−1
= (2𝑥 − 0)(3𝑥 + 5𝑥 − 2) + (2𝑥 − 1)(2 · 3𝑥
1−1
+ 5𝑥 − 0)
2
= (2)(3𝑥 + 5𝑥 − 2) + (2𝑥 − 1)(6𝑥 + 5)​​
2 2
= 6𝑥 + 10𝑥 − 4 + 12𝑥 + 10𝑥 − 6𝑥 − 5
2
= 18𝑥 + 14𝑥 − 9

3
Find the derivative of 𝑓(𝑥) = (3𝑥 − 7)(5 − 2𝑥 ).
​
Solution
𝑑 3 𝑑 3
𝑓'(𝑥) = [ 𝑑𝑥 (3𝑥 − 7)](5 − 2𝑥 ) + (3𝑥 − 7)[ 𝑑𝑥 (5 − 2𝑥 )]
1−1 3 3−1
= (3𝑥 − 0)(5 − 2𝑥 ) + (3𝑥 − 7)(0 − 3 · 2𝑥 )
3 2
= (3)(5 − 2𝑥 ) + (3𝑥 − 7)(− 6𝑥 )​ ​ ​
3 3 2
= 15 − 6𝑥 − 18𝑥 + 42𝑥
3 2
= 15 − 24𝑥 + 42𝑥
3 2
=− 24𝑥 + 42𝑥 + 15

3 69
Find the derivative of ℎ(𝑥) = ( 𝑥)(𝑥 − 3)(𝑥 − 9).
​
Solution
𝑑 3 69
ℎ'(𝑥) = 𝑑𝑥
[( 𝑥)(𝑥 − 3)(𝑥 − 9)]
In order to continue solving for this derivative, we must use the product rule for
𝑑
three functions: 𝑑𝑥
[𝑓(𝑥)𝑔(𝑥)ℎ(𝑥)] = 𝑓'(𝑥)𝑔(𝑥)ℎ(𝑥) + 𝑓(𝑥)𝑔'(𝑥)ℎ(𝑥) + 𝑓(𝑥)𝑔(𝑥)ℎ'(𝑥).
Consequently,

62
 1 1 1
𝑑 3 69 𝑑 3 69 3 𝑑 69
ℎ'(𝑥) = [ 𝑑𝑥 (𝑥 2 )](𝑥 − 3)(𝑥 − 9) + (𝑥 2 )[ 𝑑𝑥 (𝑥 − 3)](𝑥 −) + (𝑥 2 )(𝑥 − 3)[ 𝑑𝑥 (𝑥 − 9)]
​ ​
1 1 1
1 2
−1 3 69 3−1 69 3
= [( 𝑥2
)](𝑥 − 3)(𝑥 − 9) + (𝑥 2 )[(3 · 𝑥 − 0)](𝑥 − 9) + (𝑥 2 )(𝑥
69−1
− 3)[(69 · 𝑥 − 0)]
−1 1 1
1 2 3 69 2 69 3 68
= (2𝑥 )(𝑥 − 3)(𝑥 − 9) + (𝑥 2 )(3𝑥 )(𝑥 − 9) + (𝑥 2 )(𝑥 − 3)(69𝑥 )
1 1
1 3 69 2 69 3 68
=( 1 )(𝑥 − 3)(𝑥 − 9) + (𝑥 2 )(3𝑥 )(𝑥 − 9) + (𝑥 2 )(𝑥 − 3)(69𝑥 )
2
2𝑥
1 1
1 3 69 2 69 3 68
=( 1 )(𝑥 − 3)(𝑥 − 9) + (𝑥 2 )(3𝑥 )(𝑥 − 9) + (𝑥 2 )(𝑥 − 3)(69𝑥 )
2
2𝑥
3 69 5 137
(𝑥 −3)(𝑥 −9) 69 3
= 1 + 3𝑥 2 (𝑥 − 9) + 69𝑥 2
(𝑥 − 3)
2
2𝑥
5 1 137 1
3 69 69 3
(𝑥 −3)(𝑥 −9) 3𝑥 2 (𝑥 −9)2𝑥 2 69𝑥 2
(𝑥 −3)2𝑥 2
= 1 + 1 + 1

2𝑥 2 2𝑥 2 2𝑥 2
72 3 69 72 3 72 69
(𝑥 −9𝑥 −3𝑥 +27)+(6𝑥 −54𝑥 )+(138𝑥 −414𝑥 )
= 1
2
2𝑥
72 3 69 72 3 72 69
𝑥 −9𝑥 −3𝑥 +27+6𝑥 −54𝑥 +138𝑥 −414𝑥
= 1
2
2𝑥
72 69 3
145𝑥 −417𝑥 −63𝑥 +27
= 1
2
2𝑥
1
72 69 3
145𝑥 −417𝑥 −63𝑥 +27 𝑥2
= 1 · 1
2
2𝑥 𝑥2
145 139 7 1
2
145𝑥 −417𝑥 2 −63𝑥 3 +27𝑥 2
= 2𝑥
.

2 4𝑒
Find the derivative of 𝑏(𝑥) = (𝑥 − 69)(𝑥 + 3)(𝑥 + 3𝑥 − 69
).
​
Solution
In order to continue solving for this derivative, we must use the product rule for
𝑑
three functions 𝑑𝑥
[𝑓(𝑥)𝑔(𝑥)ℎ(𝑥)] = 𝑓'(𝑥)𝑔(𝑥)ℎ(𝑥) + 𝑓(𝑥)𝑔'(𝑥)ℎ(𝑥) + 𝑓(𝑥)𝑔(𝑥)ℎ'(𝑥).
Consequently,
𝑑 2 4𝑒 𝑑
𝑏'(𝑥) = [ 𝑑𝑥 (𝑥 − 69)](𝑥 + 3)(𝑥 + 3𝑥 − 69
) + (𝑥 − 69)[ 𝑑𝑥 (𝑥 + 3)]
2 4𝑒 𝑑 2 4𝑒
+ (𝑥 + 3𝑥 − 69
) + (𝑥 − 69)(𝑥 + 3)[ 𝑑𝑥 (𝑥 + 3𝑥 − 69
)]
1−1 2 4𝑒 1−1 2
= (𝑥 − 0)(𝑥 + 3)(𝑥 + 3𝑥 − 69
) + (𝑥 − 69)(𝑥 + 0)(𝑥 +
4𝑒 2−1 1−1
3𝑥 − 69
) + (𝑥 − 69)(𝑥 + 3)(2 · 𝑥 + 3𝑥 − 0)
2 4𝑒 2 4𝑒
= (1)(𝑥 + 3)(𝑥 + 3𝑥 − 69
) + (𝑥 − 69)(1)(𝑥 + 3𝑥 − 69
)+
(𝑥 − 69)(𝑥 + 3)(2𝑥 + 3)
2 4𝑒 2 4𝑒
= (𝑥 + 3)(𝑥 + 3𝑥 − 69
) + (𝑥 − 69)(𝑥 + 3𝑥 − 69
)+
(𝑥 − 69)(𝑥 + 3)(2𝑥 + 3).

Further simplification of this derivative yields:

3 2
276𝑥 −13041𝑥 +(−8𝑒−55890)𝑥+264𝑒−42849
​ ​ = 69
.

63
The Quotient Rule

In calculus, differentiation helps us and denominator separately, we would get

determine how a function changes as its input incorrect results. To see why, consider a
changes. While some functions are simple situation where a function represents a ratio of
sums or products of individual terms, others two changing quantities—such as velocity per
involve division, where one function is divided unit time, population density, or efficiency. If
by another. In such cases, we must use the both the numerator and denominator are
Quotient Rule, which provides a structured changing, the rate of change of their ratio
method for differentiating the ratio of two depends not only on how fast each is changing
differentiable functions. The Quotient Rule but also on how they interact with each other.
states that if we have two differentiable
functions as a ratio, 𝑓(𝑥) and 𝑔(𝑥), defined as The Quotient Rule accounts for:
ℎ(𝑥), then the derivative of ℎ(𝑥), denoted as ℎ'(𝑥), is
given by: ●​ The rate at which the numerator
changes while keeping the
𝑑 𝑑

​
𝑑 𝑓(𝑥)
[ 𝑔(𝑥) ] =
𝑓(𝑥) 𝑑𝑥 𝑔(𝑥)−𝑔(𝑥) 𝑑𝑥 𝑓(𝑥)
. denominator momentarily constant.
𝑑𝑥 2
[𝑔(𝑥)] ●​ The rate at which the denominator
changes while keeping the numerator
The quotient rule is necessary momentarily constant.
because differentiation does not distribute over ●​ The effect of the denominator’s size in
division in the same way it does for addition or influencing the overall rate of change.​
subtraction. If we naively tried to differentiate
𝑓(𝑥)
𝑔(𝑥)
by simply differentiating the numerator

SAMPLE PROBLEMS | Differentiation using the Quotient Rule

𝑑 9𝑥−3
Find the derivative of 𝑑𝑥
( 1−3𝑥 ).
​
Solution
𝑑 𝑑
𝑑 9𝑥−3 ( 𝑑𝑥 (9𝑥−3))(1−3𝑥)−(9𝑥−3)( 𝑑𝑥 (1−3𝑥))
𝑑𝑥
( 1−3𝑥 ) = 2
[1−3𝑥]
1−1 1−1
(9𝑥 −0)(1−3𝑥)−(9𝑥−3)(0−3𝑥 )
= 2
[1−3𝑥]
(9)(1−3𝑥)−(9𝑥−3)(−3)
= 2
[1−3𝑥]
9−27𝑥+27𝑥−9
= 2
[1−3𝑥]
0
= 2
[1−3𝑥]
=0

2
𝑑 𝑥
Find the derivative of 𝑑𝑥
[ 𝑥−8 ].
​
Solution
2 𝑑 2 2 𝑑
𝑑 𝑥 ( 𝑑𝑥 𝑥 )(𝑥−8)−𝑥 ( 𝑑𝑥 (𝑥−8))
𝑑𝑥
[ 𝑥−8 ] = 2
[𝑥−8]

64
 2−1 2 1−1
(2·𝑥 )(𝑥−8)−𝑥 (𝑥 −0)
= 2
[𝑥−8]
2
2𝑥(𝑥−8)−𝑥
= 2
[𝑥−8]
2 2
2𝑥 −16𝑥−𝑥
= 2
[𝑥−8]
2
𝑥 −16𝑥
= 2
[𝑥−8]

𝑑 𝑥
Find the derivative of 𝑑𝑥
( 10𝑥−5 ).
​
Solution
𝑑 𝑑
𝑑 𝑥 ( 𝑑𝑥 𝑥)(10𝑥−5)−(𝑥)( 𝑑𝑥 (10𝑥−5))
𝑑𝑥
( 10𝑥−5 ) = 2
[3𝑥−5]
1−1 1−1
𝑥 (10𝑥−5)−(𝑥)(10𝑥 −0)
= 2
[3𝑥−5]
(1)(10𝑥−5)−(𝑥)(10)
= 2 ​ ​ ​ ​ ​ ​
[3𝑥−5]
10𝑥−5−10𝑥
= 2 ​
[3𝑥−5]
−5
= 2 ​ ​
[3𝑥−5]

2
𝑑 𝑥 +4
Find the derivative of 𝑑𝑥
( 2 ).
𝑥 −4
​
Solution
2 𝑑 2 2 2 𝑑 2
𝑑 𝑥 +4 ( 𝑑𝑥 (𝑥 +4))(𝑥 −4)−(𝑥 +4)( 𝑑𝑥 (𝑥 −4))
𝑑𝑥
( 2 )= 2 2
𝑥 −4 [𝑥 −4]
2−1 2 2 1−1
(2·𝑥 +0)(𝑥 −4)−(𝑥 +4)(2·𝑥 −0)
= 2 2
[𝑥 −4]
2 2
(2𝑥)(𝑥 −4)−(𝑥 +4)(2𝑥)
= 2 2
[𝑥 −4]
3 3
2𝑥 −8𝑥−2𝑥 −8𝑥
= 2 2
[𝑥 −4]
−16𝑥
= 2 2
[𝑥 −4]

4
𝑑 (6𝑥 −1)
Find the derivative of 𝑑𝑥
[ 3𝑥−2
].
​
Solution
4 𝑑 4 4 𝑑
𝑑 (6𝑥 −1) ( 𝑑𝑥 (6𝑥 −1))(3𝑥−2)−(6𝑥 −1)( 𝑑𝑥 (3𝑥−2))
𝑑𝑥
[ 3𝑥−2
]= 2
[3𝑥−2]
4−1 4 1−1
(4·6𝑥 −0)(3𝑥−2)−(6𝑥 −1)(3𝑥 −0)
= 2
[3𝑥−2]
3 4
(24𝑥 )(3𝑥−2)−(6𝑥 −1)(3)
= 2
[3𝑥−2]
4 3 4
72𝑥 −48𝑥 −18𝑥 +3
= 2
[3𝑥−2]

65
 4 3
54𝑥 −48𝑥 +3
= 2
[3𝑥−2]

Derivatives of Natural Exponential Functions

The natural exponential function is one 𝑥

nature makes 𝑒 the only function in calculus
of the most fundamental functions in calculus
that remains unchanged under differentiation.
and mathematical modeling. It is defined as 𝑥
𝑥 The natural exponential function 𝑒 is one of
𝑓(𝑥) = 𝑒 , where 𝑒 (Euler’s number) is an
the most important functions in mathematics
irrational constant approximately equal to
because:
2.71828. The key characteristic of the natural
exponential function is that its rate of growth is ●​ It is the only function that is its own
proportional to its current value, which is a derivative.​
defining property of many natural processes.
The natural exponential function is unique ●​ It plays a fundamental role in growth
because its derivative is the function itself: and decay models.​
𝑑 𝑥 𝑥
​ ​ 𝑑𝑥
𝑒 =𝑒 ●​ It appears in many scientific, financial,
and engineering applications.​
This means that if we differentiate
exe^xex any number of times, it will always ●​ It forms the basis for solving
return to the original function. self-replicating differential equations.

SAMPLE PROBLEMS | Derivatives of Natural Exponential Functions

3𝑥+1
Find the derivative of 𝑧(𝑥) = 𝑒 .
​
Solution
3𝑥+1 1−1
𝑧'(𝑥) = 𝑒 · (3𝑥 + 0)
3𝑥+1
= 3𝑒

𝑒𝑥
Find the derivative of 𝑗(𝑥) = 𝑒 .
​
Solution
𝑒𝑥 1−1
𝑗'(𝑥) = 𝑒 · 𝑒𝑥
𝑒𝑥
=𝑒 ·𝑒
2𝑒𝑥
=𝑒

5𝑥
Find the derivative of 𝑓(𝑥) = 69𝑒 .

66
 ​
Solution
5𝑥 1−1
𝑓'(𝑥) = 69𝑒 · 5𝑥
5𝑥
= 69𝑒 ·5
5𝑥
= 345𝑒

10𝑥 77𝑥+3
Find the derivative of ℎ(𝑥) = 𝑒 +𝑒 .
​
Solution
10𝑥 1−1 77𝑥+3 1−1
ℎ'(𝑥) = 𝑒 · 10𝑥 +𝑒 · (77𝑥 + 0)
10𝑥 77𝑥+3
=𝑒 · 10 + 𝑒 · 77
10𝑥 77𝑥+3
= 10𝑒 + 77𝑒

𝑥+9
Find the derivative of 𝑣(𝑥) = 12𝑥𝑒 .
​
Solution
𝑑 𝑥+9 𝑑 𝑥+9
𝑣'(𝑥) = ( 𝑑𝑥 (12𝑥))𝑒 + 12𝑥( 𝑑𝑥 (𝑒 ))
1−1 𝑥+9 𝑥+9 1−1
= (12𝑥 )𝑒 + 12𝑥(𝑒 ·𝑥 )
𝑥+9 𝑥+9
= 12𝑒 + 12𝑥𝑒
𝑥+9
= 12𝑒 (1 + 𝑥)

Derivatives of General Exponential Functions

Exponential functions are fundamental The derivative of an exponential function

in mathematics, physics, engineering, and 𝑥
𝑓(𝑥) = 𝑎 follows the formula:
economics. They describe rapid growth and
decay processes, such as population growth, 𝑑 𝑥 𝑥
radioactive decay, and compound interest. The ​ ​ 𝑑𝑥
(𝑎 ) = 𝑎 𝑙𝑛(𝑎)
general form of an exponential function is
𝑥
𝑓(𝑥) = 𝑎 , where a is a positive constant base where 𝑙𝑛(𝑎) is the natural logarithm of the
(i.e., a>0 and a ≠ 1). base 𝑎.
This formula tells us that the derivative of an
The derivative of an exponential exponential function is proportional to the
function is unique because the function itself function itself, with the proportionality constant
contains the variable in the exponent, unlike being 𝑙𝑛(𝑎).
polynomial functions where the exponent is a
constant.

67
SAMPLE PROBLEMS | Derivatives of General Exponential Functions

𝑥
Find the derivative of 𝑧(𝑥) = 69 .
​
Solution
𝑥 1−1
​ 𝑧'(𝑥) = 69 𝑙𝑛(69) · 𝑥
𝑥
= 69 𝑙𝑛(69)

𝑥−1
Find the derivative of 𝑓(𝑥) = 9 .
​
Solution
𝑥−1 1−1
𝑓'(𝑥) = 9 𝑙𝑛(9) · 𝑥
𝑥−1
=9 𝑙𝑛(9)

12𝑥
Find the derivative of 𝑝(𝑥) = π .
​
Solution
12𝑥 1−1
𝑝'(𝑥) = π 𝑙𝑛(π) · 12𝑥
12𝑥
= 12π 𝑙𝑛(π)

𝑥 𝑥−9
Find the derivative of 𝑔(𝑥) = 5 · 23 .
​
Solution
𝑥 1−1 𝑥−9 𝑥 𝑥−9 1−1
𝑔'(𝑥) = 5 𝑙𝑛(5) · 𝑥 · 23 + 5 · 23 𝑙𝑛(23) · (𝑥 − 0)
𝑥 𝑥−9 𝑥 𝑥−9
= 5 𝑙𝑛(5) · 23 + 5 · 23 𝑙𝑛(23)
𝑥 𝑥−9
= 5 · 23 (𝑙𝑛(5) + 𝑙𝑛(23))

𝑥 𝑥+9
Find the derivative of 𝑜(𝑥) = 30 + 7 .
​
Solution
𝑥 1−1 𝑥+9 1−1
𝑜'(𝑥) = 30 𝑙𝑛(30) · 𝑥 +7 𝑙𝑛(7) · (𝑥 + 0)
𝑥 𝑥+9
= 30 𝑙𝑛(30) + 7 𝑙𝑛(7)

68
Derivatives of Trigonometric Functions

Trigonometric functions play a demonstrating their rapid growth near vertical

fundamental role in calculus, modeling asymptotes. The secant function 𝑠𝑒𝑐(𝑥) has a
oscillatory behavior and wave phenomena. derivative of 𝑠𝑒𝑐(𝑥), while the cosecant
Their derivatives describe the rate of change of function 𝑐𝑠𝑐(𝑥) follows with − 𝑐𝑠𝑐(𝑥)𝑐𝑜𝑡(𝑥).
these functions, making them essential in These derivatives help analyze wave motion,
physics, engineering, and other sciences. The electrical circuits, economic cycles, and
derivative of 𝑠𝑖𝑛(𝑥) is 𝑐𝑜𝑠(𝑥), while the biological rhythms, among other applications.
derivative of 𝑐𝑜𝑠(𝑥) is − 𝑠𝑖𝑛(𝑥), reflecting their
cyclic nature. The derivative of 𝑡𝑎𝑛(𝑥) is
2 2
𝑠𝑒𝑐 (𝑥), and for 𝑐𝑜𝑡(𝑥), it is − 𝑐𝑠𝑐 (𝑥),

The derivatives of the corresponding trigonometric functions can be summarized in this following
table:

FUNCTION NOTATION

𝑠𝑖𝑛(𝑥) 𝑐𝑜𝑠(𝑥)

𝑐𝑜𝑠(𝑥) − 𝑠𝑖𝑛(𝑥)

𝑡𝑎𝑛(𝑥) 2
𝑠𝑒𝑐 (𝑥)

𝑐𝑜𝑡(𝑥) 2
− 𝑐𝑠𝑐 (𝑥)

𝑠𝑒𝑐(𝑥) 𝑠𝑒𝑐(𝑥)𝑡𝑎𝑛(𝑥)

𝑐𝑠𝑐(𝑥) − 𝑐𝑠𝑐(𝑥)𝑐𝑜𝑡(𝑥)

SAMPLE PROBLEMS | Derivatives of Trigonometric Functions

Find the derivative of 𝑓(𝑥) = 𝑠𝑒𝑐(𝑥)𝑐𝑠𝑐(𝑥)

​
Solution
𝑑 𝑑
𝑓'(𝑥) = ( 𝑑𝑥 (𝑠𝑒𝑐(𝑥)))𝑐𝑠𝑐(𝑥) − 𝑠𝑒𝑐(𝑥)( 𝑑𝑥 (𝑐𝑠𝑐(𝑥)))
= 𝑠𝑒𝑐(𝑥)𝑡𝑎𝑛(𝑥)𝑐𝑠𝑐(𝑥) − 𝑠𝑒𝑐(𝑥)𝑐𝑠𝑐(𝑥)𝑐𝑜𝑡(𝑥)
1 𝑠𝑖𝑛(𝑥) 1 1 1 𝑐𝑜𝑠(𝑥)
= 𝑐𝑜𝑠(𝑥)
· 𝑐𝑜𝑠(𝑥)
· 𝑠𝑖𝑛(𝑥)
− 𝑐𝑜𝑠(𝑥)
· 𝑠𝑖𝑛(𝑥)
· 𝑠𝑖𝑛(𝑥)
1 1
= 2 − 2
𝑐𝑜𝑠 (𝑥) 𝑠𝑖𝑛 (𝑥)
2 2
= 𝑠𝑒𝑐 (𝑥) − 𝑐𝑠𝑐 (𝑥)

Find the derivative of 𝑘(𝑥) = 𝑡𝑎𝑛(𝑥)𝑠𝑒𝑐(𝑥).

​
Solution

69
 𝑑 𝑑
𝑘'(𝑥) = ( 𝑑𝑥 (𝑡𝑎𝑛(𝑥)))𝑠𝑒𝑐(𝑥) + 𝑡𝑎𝑛(𝑥)( 𝑑𝑥 (𝑠𝑒𝑐(𝑥)))
2
= 𝑠𝑒𝑐 (𝑥)𝑠𝑒𝑐(𝑥) + 𝑡𝑎𝑛(𝑥)𝑠𝑒𝑐(𝑥)𝑡𝑎𝑛(𝑥)
2
3 𝑡𝑎𝑛 (𝑥)
= 𝑠𝑒𝑐 (𝑥) + 𝑠𝑒𝑐(𝑥)
2
𝑠𝑖𝑛 (𝑥)
3 2
𝑐𝑜𝑠 (𝑥)
= 𝑠𝑒𝑐 (𝑥) + 1
𝑐𝑜𝑠(𝑥)
2
3 𝑠𝑖𝑛 (𝑥)
= 𝑠𝑒𝑐 (𝑥) + 𝑐𝑜𝑠(𝑥)
3
= 𝑠𝑒𝑐 (𝑥) + 𝑡𝑎𝑛(𝑥)𝑠𝑖𝑛(𝑥)

𝑡𝑎𝑛(𝑥)
Find the derivative of 𝑣(𝑥) = 𝑠𝑒𝑐(𝑥)
.
​
Solution
𝑑 𝑑
( 𝑑𝑥 (𝑡𝑎𝑛(𝑥)))𝑠𝑒𝑐(𝑥)−𝑡𝑎𝑛(𝑥)( 𝑑𝑥 (𝑠𝑒𝑐(𝑥)))
𝑣'(𝑥) = 2
[𝑠𝑒𝑐(𝑥)]
2
𝑠𝑒𝑐 (𝑥)𝑠𝑒𝑐(𝑥)−𝑡𝑎𝑛(𝑥)𝑠𝑒𝑐(𝑥)𝑡𝑎𝑛(𝑥)
= 2
[𝑠𝑒𝑐(𝑥)]
2
3 𝑡𝑎𝑛 (𝑥)
𝑠𝑒𝑐 (𝑥)− 𝑠𝑒𝑐(𝑥)
= 2
𝑠𝑒𝑐 (𝑥)
2
𝑠𝑖𝑛 (𝑥)
3 2
𝑐𝑜𝑠 (𝑥)
𝑠𝑒𝑐 (𝑥)− 1

= 2
𝑐𝑜𝑠(𝑥)

𝑠𝑒𝑐 (𝑥)
2
3 𝑠𝑖𝑛 (𝑥)
𝑠𝑒𝑐 (𝑥)− 𝑐𝑜𝑠(𝑥)
= 2
𝑠𝑒𝑐 (𝑥)
3
𝑠𝑒𝑐 (𝑥)−𝑡𝑎𝑛(𝑥)𝑠𝑖𝑛(𝑥)
= 2
𝑠𝑒𝑐 (𝑥)

Find the derivative of 𝑗(𝑥) = 𝑐𝑜𝑡(𝑥)𝑐𝑠𝑐(𝑥)

.
Solution
𝑑 𝑑
𝑗'(𝑥) = ( 𝑑𝑥 (𝑐𝑜𝑡(𝑥)))𝑐𝑠𝑐(𝑥) + 𝑐𝑜𝑡(𝑥)( 𝑑𝑥 (𝑐𝑠𝑐(𝑥)))
2
=− 𝑐𝑠𝑐 (𝑥)𝑐𝑠𝑐(𝑥) − 𝑐𝑜𝑡(𝑥)𝑐𝑠𝑐(𝑥)𝑐𝑜𝑡(𝑥)
3 2
=− 𝑐𝑠𝑐 (𝑥) − 𝑐𝑜𝑡 (𝑥)𝑐𝑠𝑐(𝑥)

1
Find the derivative of 𝑘(𝑥) = [(69𝑠𝑖𝑛(𝑥))(𝑐𝑜𝑠(𝑥))( 3 𝑡𝑎𝑛(𝑥))].
​
Solution
𝑑
𝑘'(𝑥) = 23 · 𝑑𝑥
[𝑠𝑖𝑛(𝑥)𝑐𝑜𝑠(𝑥)𝑡𝑎𝑛(𝑥)]
​ In order to continue solving for this derivative, we must use the product rule for
𝑑
three functions 𝑑𝑥
[𝑓(𝑥)𝑔(𝑥)ℎ(𝑥)] = 𝑓'(𝑥)𝑔(𝑥)ℎ(𝑥) + 𝑓(𝑥)𝑔'(𝑥)ℎ(𝑥) + 𝑓(𝑥)𝑔(𝑥)ℎ'(𝑥).
Consequently,
𝑑 𝑑 𝑑
𝑘'(𝑥) = 23[( 𝑑𝑥 (𝑠𝑖𝑛(𝑥)))𝑐𝑜𝑠(𝑥)𝑡𝑎𝑛(𝑥) + 𝑠𝑖𝑛(𝑥)( 𝑑𝑥 (𝑐𝑜𝑠(𝑥)))𝑡𝑎𝑛(𝑥) + 𝑐𝑜𝑠(𝑥)𝑠𝑖𝑛(𝑥)( 𝑑𝑥 (𝑡𝑎𝑛(𝑥)))]

70
 2
= 23[− 𝑠𝑖𝑛(𝑥)𝑠𝑖𝑛(𝑥)𝑡𝑎𝑛(𝑥) + 𝑐𝑜𝑠(𝑥)𝑐𝑜𝑠(𝑥)𝑡𝑎𝑛(𝑥) + 𝑐𝑜𝑠(𝑥)𝑠𝑖𝑛(𝑥)𝑠𝑒𝑐 (𝑥)]
2 2 2
=− 23𝑠𝑖𝑛 (𝑥)𝑡𝑎𝑛(𝑥) + 23𝑐𝑜𝑠 (𝑥)𝑡𝑎𝑛(𝑥) + 23𝑐𝑜𝑠(𝑥)𝑠𝑒𝑐 (𝑥)𝑠𝑖𝑛(𝑥)
2 2 2
=− 23[(𝑠𝑖𝑛 (𝑥) − 𝑐𝑜𝑠 (𝑥))𝑡𝑎𝑛(𝑥) − 𝑐𝑜𝑠(𝑥)𝑠𝑒𝑐 (𝑥)𝑠𝑖𝑛(𝑥)]
=− 23[− 𝑐𝑜𝑠(2𝑥)𝑡𝑎𝑛(𝑥) − 𝑠𝑒𝑐(𝑥)𝑠𝑖𝑛(𝑥)]
=− 23[− 𝑐𝑜𝑠(2𝑥)𝑡𝑎𝑛(𝑥) − 𝑡𝑎𝑛(𝑥)]
= 23𝑡𝑎𝑛(𝑥)[𝑐𝑜𝑠(2𝑥) − 1]

Derivatives of Logarithmic Functions

Logarithmic functions are 𝑎 which must be raised to obtain 𝑥. The

fundamental in calculus due to their ability to most commonly used logarithmic function is
transform multiplicative relationships into the natural logarithm, which has the Euler’s
additive ones, making differentiation easier. number as its base: 𝑙𝑛(𝑥) = 𝑙𝑜𝑔𝑒𝑥, where
Their derivatives play a crucial role in many
𝑒 ≈ 2. 71828. The most fundamental
mathematical applications, including
derivative in logarithmic calculus is:
exponential growth and decay, information
theory, and complex function differentiation. 𝑑 1
𝑙𝑛(𝑥) = , 𝑥 > 0.
A logarithmic function is the inverse of an 𝑑𝑥 𝑥

exponential function. In general, it takes the

form 𝑓(𝑥) = 𝑙𝑜𝑔𝑎𝑥, where 𝑎 is the base of For a logarithmic function with an arbitrary
base 𝑎:
the logarithm, 𝑥 is the input value (must be
positive), and 𝑙𝑜𝑔𝑎𝑥 is the exponent to which 𝑑 1
𝑑𝑥
𝑙𝑜𝑔𝑎(𝑥) = 𝑥𝑙𝑛(𝑎)
, 𝑥 > 0.

SAMPLE PROBLEMS | Derivatives of Logarithmic Functions

Find the derivative of ℎ(𝑥) = 𝑙𝑜𝑔2𝑥.

Solution
1
ℎ'(𝑥) = 𝑥𝑙𝑛2

Find the derivative of 𝑙(𝑥) = 𝑙𝑜𝑔69 𝑥.

​
Solution
1
𝑙'(𝑥) = 𝑥𝑙𝑛(69)

Find the derivative of 𝑎(𝑥) = (𝑙𝑛(𝑥))(𝑙𝑜𝑔9𝑥)

​
Solution
1 1
𝑎'(𝑥) = 𝑥
𝑙𝑜𝑔9𝑥 + 𝑥𝑙𝑛(9)
𝑙𝑛(𝑥)

71
 3 2
Find the derivative of 𝑑(𝑥) = 2𝑥 𝑙𝑜𝑔(𝑥 + 69𝑥 − 34).
​
Solution
2 2 3 1
𝑑'(𝑥) = 6𝑥 𝑙𝑜𝑔(𝑥 + 69𝑥 − 34) + 2𝑥 · 2
(𝑥 +69𝑥−34)𝑙𝑛(10)
3
2 2 2𝑥
= 6𝑥 𝑙𝑜𝑔(𝑥 + 69𝑥 − 34) + 2
(𝑥 +69𝑥−34)𝑙𝑛(10)

𝑥
Find the derivative of 𝑓(𝑥) = 𝑙𝑛(𝑥)
.
​
Solution
𝑑 𝑑
( 𝑑𝑥 𝑥)𝑙𝑛(𝑥)−𝑥·( 𝑑𝑥 𝑙𝑛(𝑥))
𝑓'(𝑥) = 2
[𝑙𝑛(𝑥)]

1−1 1
𝑙𝑛𝑥·(𝑥 )−𝑥·( 𝑥 )
​ = 2
[𝑙𝑛(𝑥)]

0 𝑥
𝑙𝑛𝑥·(𝑥 )−( 𝑥 )
= 2
[𝑙𝑛(𝑥)]

𝑙𝑛(𝑥)·(1)−(1)
= 2
[𝑙𝑛(𝑥)]

𝑙𝑛(𝑥)−1
​ = 2
[𝑙𝑛(𝑥)]

Derivatives of Inverse Trigonometric Functions

Inverse trigonometric functions are fundamental in calculus as they allow us to reverse

standard trigonometric functions and find angles given trigonometric ratios. Their derivatives are
essential in solving integrals, analyzing motion, and working with implicit differentiation.

The six inverse trigonometric functions correspond to the six basic trigonometric functions:
FUNCTION INVERSE FUNCTION NOTATION

Sine Arcsine −1
𝑠𝑖𝑛 (𝑥) 𝑜𝑟 𝑎𝑟𝑐𝑠𝑖𝑛(𝑥)

Cosine Arcosine −1
𝑐𝑜𝑠 (𝑥) 𝑜𝑟 𝑎𝑟𝑐𝑐𝑜𝑠(𝑥)

Tangent Arctangent −1
𝑡𝑎𝑛 (𝑥) 𝑜𝑟 𝑎𝑟𝑐𝑡𝑎𝑛(𝑥)

Cotangent Arcotangent −1
𝑐𝑜𝑡 (𝑥) 𝑜𝑟 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥)

Secant Arcsecant −1
𝑠𝑒𝑐 (𝑥) 𝑜𝑟 𝑎𝑟𝑐𝑠𝑒𝑐(𝑥)

72
 Cosecant Arcosecant −1
𝑐𝑠𝑐 (𝑥) 𝑜𝑟 𝑎𝑟𝑐𝑐𝑠𝑐(𝑥)

The six inverse trigonometric functions also correspond to each of their derivatives:

FUNCTION DERIVATIVE

−1 𝑑 −1 1
𝑠𝑖𝑛 (𝑥) 𝑜𝑟 𝑎𝑟𝑐𝑠𝑖𝑛(𝑥) 𝑑𝑥
[𝑠𝑖𝑛 (𝑥)] = 2
, 𝑥 ≠± 1
1−𝑥

−1 𝑑 −1 −1
𝑐𝑜𝑠 (𝑥) 𝑜𝑟 𝑎𝑟𝑐𝑐𝑜𝑠(𝑥) 𝑑𝑥
[𝑐𝑜𝑠 (𝑥)] = 2
, 𝑥 ≠± 1
1−𝑥

−1 𝑑 −1 1
𝑡𝑎𝑛 (𝑥) 𝑜𝑟 𝑎𝑟𝑐𝑡𝑎𝑛(𝑥) 𝑑𝑥
[𝑡𝑎𝑛 (𝑥)] = 2
1+𝑥

−1 𝑑 −1 −1
𝑐𝑜𝑡 (𝑥) 𝑜𝑟 𝑎𝑟𝑐𝑐𝑜𝑡(𝑥) 𝑑𝑥
[𝑐𝑜𝑡 (𝑥)] = 2
1+𝑥

−1 𝑑 −1 1
𝑠𝑒𝑐 (𝑥) 𝑜𝑟 𝑎𝑟𝑐𝑠𝑒𝑐(𝑥) 𝑑𝑥
[𝑠𝑒𝑐 (𝑥)] = 2
, 𝑥 ≠± 1, 0
|𝑥| 𝑥 −1

−1 𝑑 −1 −1
𝑐𝑠𝑐 (𝑥) 𝑜𝑟 𝑎𝑟𝑐𝑐𝑠𝑐(𝑥) 𝑑𝑥
[𝑐𝑠𝑐 (𝑥)] = 2
, 𝑥 ≠± 1, 0
|𝑥| 𝑥 −1

SAMPLE PROBLEMS | Derivatives of Inverse Trigonometric Functions

Find the derivative of 𝑡(𝑥) = 𝑠𝑖𝑛(𝑥)𝑎𝑟𝑐𝑡𝑎𝑛(𝑥 − 1).

Solution
𝑑 𝑑
𝑡'(𝑥) = ( 𝑑𝑥 (𝑠𝑖𝑛(𝑥)))𝑎𝑟𝑐𝑡𝑎𝑛(𝑥 − 1) + 𝑠𝑖𝑛(𝑥)( 𝑑𝑥 (𝑎𝑟𝑐𝑡𝑎𝑛(𝑥)))
1
= 𝑐𝑜𝑠(𝑥)𝑎𝑟𝑐𝑡𝑎𝑛(𝑥 − 1) + 𝑠𝑖𝑛(𝑥) 2
𝑥 +1
𝑠𝑖𝑛(𝑥)
= 𝑐𝑜𝑠(𝑥)𝑎𝑟𝑐𝑡𝑎𝑛(𝑥 − 1) + 2
𝑥 +1

Find the derivative of 𝑚(𝑥) = 𝑎𝑟𝑐𝑠𝑖𝑛(𝑥 − 9).

​
Solution
1
𝑚'(𝑥) = 2
1−(𝑥−9)
1
= 2
1−(𝑥 −18𝑥+81)
1
= 2
1−𝑥 +18𝑥−81
1
= 2
−𝑥 +18𝑥−80
2
1 −𝑥 +18𝑥−80
= 2
· 2
−𝑥 +18𝑥−80 −𝑥 +18𝑥−80

73
 2
−𝑥 +18𝑥−80
= 2
−𝑥 +18𝑥−80

Find the derivative of 𝑟(𝑥) = 𝑎𝑟𝑐𝑡𝑎𝑛(7 + 23𝑥) · 𝑎𝑟𝑐𝑠𝑖𝑛(𝑥 − 9).

​
Solution
𝑑 𝑑
𝑟'(𝑥) = ( 𝑑𝑥 (𝑎𝑟𝑐𝑡𝑎𝑛(7 + 23𝑥)))𝑎𝑟𝑐𝑠𝑖𝑛(𝑥 − 9) + 𝑎𝑟𝑐𝑡𝑎𝑛(7 + 23𝑥)( 𝑑𝑥 (𝑎𝑟𝑐𝑠𝑖𝑛(𝑥 − 9)))
1 1
= 2 𝑎𝑟𝑐𝑠𝑖𝑛(𝑥 − 9) + 𝑎𝑟𝑐𝑡𝑎𝑛(7 + 23𝑥)( 2
)
1+(7+23𝑥) −𝑥 +18𝑥−80
𝑎𝑟𝑐𝑠𝑖𝑛(𝑥−9) 𝑎𝑟𝑐𝑡𝑎𝑛(7+23𝑥)
= 2 +( 2
)
1+(7+23𝑥) −𝑥 +18𝑥−80
2
𝑎𝑟𝑐𝑠𝑖𝑛(𝑥−9) 𝑎𝑟𝑐𝑡𝑎𝑛(7+23𝑥) −𝑥 +18𝑥−80
= 2 +( 2
· 2
)
1+(7+23𝑥) −𝑥 +18𝑥−80 −𝑥 +18𝑥−80
2
𝑎𝑟𝑐𝑠𝑖𝑛(𝑥−9) 𝑎𝑟𝑐𝑡𝑎𝑛(7+23𝑥) −𝑥 +18𝑥−80
= 2 +( 2 )
1+(7+23𝑥) −𝑥 +18𝑥−80

Find the derivative of 𝑓(𝑥) = 𝑎𝑟𝑐𝑐𝑠𝑐(𝑥) + 𝑎𝑟𝑐𝑠𝑖𝑛(𝑥)𝑎𝑟𝑐𝑐𝑜𝑠(𝑥 − 3).

​
Solution
𝑑 𝑑 𝑑
𝑓'(𝑥) = ( 𝑑𝑥 (𝑎𝑟𝑐𝑐𝑠𝑐(𝑥))) + ( 𝑑𝑥 (𝑎𝑟𝑐𝑠𝑖𝑛(𝑥)))𝑎𝑟𝑐𝑐𝑜𝑠(𝑥 − 3) + 𝑎𝑟𝑐𝑠𝑖𝑛(𝑥)( 𝑑𝑥 (𝑎𝑟𝑐𝑐𝑜𝑠(𝑥 − 3)))
−1 1 ⎡ −1 ⎤
= 2
+ 2
(𝑎𝑟𝑐𝑐𝑜𝑠(𝑥 − 3)) + 𝑎𝑟𝑐𝑠𝑖𝑛(𝑥)⎢ 2 ⎥
|𝑥| 𝑥 −1 1−𝑥 ⎣ 1−(𝑥−3) ⎦
−1 𝑎𝑟𝑐𝑐𝑜𝑠(𝑥−3) 𝑎𝑟𝑐𝑠𝑖𝑛(𝑥)
= 2
+ 2
− 2
|𝑥| 𝑥 −1 1−𝑥 1−(𝑥 −6𝑥+9)
−1 𝑎𝑟𝑐𝑐𝑜𝑠(𝑥−3) 𝑎𝑟𝑐𝑠𝑖𝑛(𝑥)
= 2
+ 2
− 2
|𝑥| 𝑥 −1 1−𝑥 1−𝑥 +6𝑥−9
2 2 2
−1 𝑥 −1 𝑎𝑟𝑐𝑐𝑜𝑠(𝑥−3) 1−𝑥 𝑎𝑟𝑐𝑠𝑖𝑛(𝑥) −𝑥 +6𝑥−8
= 2
· 2
+ 2
· 2
− 2
· 2
|𝑥| 𝑥 −1 𝑥 −1 1−𝑥 1−𝑥 −𝑥 +6𝑥−8 −𝑥 +6𝑥−8
2 2 2
− 𝑥 −1 𝑎𝑟𝑐𝑐𝑜𝑠(𝑥−3) 1−𝑥 𝑎𝑟𝑐𝑠𝑖𝑛(𝑥) −𝑥 +6𝑥−8
= 2 + 2 − 2
|𝑥|(𝑥 −1) (1−𝑥 ) (−𝑥 +6𝑥−8)

𝑎𝑟𝑐𝑡𝑎𝑛(3𝑥)
Find the derivative of 𝑣(𝑥) = 𝑎𝑟𝑐𝑐𝑜𝑡(6𝑥)
.
​
Solution
𝑑 𝑑
( 𝑑𝑥 (𝑎𝑟𝑐𝑡𝑎𝑛(3𝑥)))𝑎𝑟𝑐𝑐𝑜𝑡(6𝑥)−𝑎𝑟𝑐𝑡𝑎𝑛(3𝑥)( 𝑑𝑥 (𝑎𝑟𝑐𝑐𝑜𝑡(6𝑥)))
𝑣'(𝑥) = 2
[𝑎𝑟𝑐𝑐𝑜𝑡(6𝑥)]

·3·𝑎𝑟𝑐𝑐𝑜𝑡(6𝑥)−𝑎𝑟𝑐𝑡𝑎𝑛(3𝑥)⎡⎢ ·6⎤⎥
1 −1
2 2
(3𝑥) +1 ⎣ (6𝑥) +1 ⎦
= 2
[𝑎𝑟𝑐𝑐𝑜𝑡(6𝑥)]
3𝑎𝑟𝑐𝑐𝑜𝑡(6𝑥) 6𝑎𝑟𝑐𝑡𝑎𝑛(3𝑥)
+ 2 2
9𝑥 +1
2 2
36𝑥 +1 (9𝑥 +1)(36𝑥 +1)
= 2 · 2 2
𝑎𝑟𝑐𝑐𝑜𝑡 (6𝑥) (9𝑥 +1)(36𝑥 +1)
2 2
3(36𝑥 +1)𝑎𝑟𝑐𝑐𝑜𝑡(6𝑥)+6(9𝑥 +1)𝑎𝑟𝑐𝑡𝑎𝑛(3𝑥)
= 2 2 2
(9𝑥 +1)(36𝑥 +1)𝑎𝑟𝑐𝑐𝑜𝑡 (6𝑥)

74